{"id":4907,"date":"2026-03-09T22:00:46","date_gmt":"2026-03-09T22:00:46","guid":{"rendered":"https:\/\/mrenglishkj.com\/?p=4907"},"modified":"2026-03-19T16:31:11","modified_gmt":"2026-03-19T16:31:11","slug":"sat-test-6-module-1st-math-problem-solution","status":"publish","type":"post","link":"https:\/\/us.mrenglishkj.com\/sat\/sat-test-6-module-1st-math-problem-solution\/","title":{"rendered":"SAT Math Test 6 Free Module 1st Question &amp; Hack"},"content":{"rendered":"\n<h2 class=\"wp-block-heading\">Try the SAT Math Test Module 1st with Real Exam Questions and Simple Detailed Solutions<\/h2>\n\n\n\n<p>How much have you learned so far? Are you progressing by taking the SAT Tests and also learning all Math problems by understanding solutions. This is the 6th Math test. Like the other exams, it has the same format and all the necessary features for you to become a master in math. You just take the SAT Test Module First to practice your skills. The best part is that you practice within the time limit, and there are explanations of the correct answers and tips and tricks to get a perfect score on the SAT. You will find Math easy after step-by-step explanations.<\/p>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<h3 class=\"wp-block-heading\">ABOUT THE SAT MODULES<\/h3>\n\n\n\n<p>The SAT is divided into four modules. There are two categories with each split into two modules. The first category is &#8220;Reading and Writing&#8221; with two modules. The second category is &#8220;Math&#8221; with two modules. The one, you will do below is SAT Math Practice Test Module 1st.<\/p>\n\n\n\n<p>The first module has questions ranging from easy to difficult, but the second module only contains difficult questions. If you want to take some other SATs, visit the links below.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/us.mrenglishkj.com\/sat\/category\/sat-english\/module-1st\/\" target=\"_blank\" rel=\"noopener\" title=\"\">1st Module of SAT Reading And Writing Practice Tests<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/us.mrenglishkj.com\/sat\/category\/sat-english\/module-2nd\/\" target=\"_blank\" rel=\"noopener\" title=\"\">2nd Module of SAT Reading And Writing Practice Tests<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/us.mrenglishkj.com\/sat\/category\/sat-math\/1st-module\/\" target=\"_blank\" rel=\"noopener\" title=\"\">1st Module of SAT Math Practice Tests<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/us.mrenglishkj.com\/sat\/category\/sat-math\/2nd-module\/\" target=\"_blank\" rel=\"noopener\" title=\"\">2nd Module of SAT Math Practice Tests<\/a><\/li>\n<\/ul>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<h3 class=\"wp-block-heading\">THE SAT MATH MODULE 1ST<\/h3>\n\n\n\n<p>The first module of Math in SAT contains four segments: Algebra, Advanced Math, Problem-Solving and Data Analysis, and Geometry and Trigonometry. The questions in Module 1st are from easy to difficult. In a real SAT exam, you must answer 22 questions within 35 minutes. We have provided you with the same in this Practice Test.<\/p>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Instructions for the SAT Real-Time Exam<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Go Back-and-Forth:<\/strong> You will see an arrow on the right or left corner of the slide, click to move forward or backward.<\/li>\n\n\n\n<li><strong>Interaction:<\/strong> You will see a press button at the top right corner that tells you there are some interactive components in the slide. Click the press button to find out.<\/li>\n\n\n\n<li><strong>Timer: <\/strong>On the top of the slide, you will see the timer, we have divided the time based on the average of the module 1st. (The 35 minutes are equally divided into 22 questions&#8217; time.) It is best to note the time before and after finishing the practice test to measure, &#8220;Was it within 35 minutes or not?&#8221;<\/li>\n\n\n\n<li><strong>Mute:<\/strong> You can click on the speaker button to mute the audio.<\/li>\n\n\n\n<li><strong>Image:<\/strong> You can click on a graph, table, or other image to expand it and view it in full screen.<\/li>\n\n\n\n<li><strong>Mobile:<\/strong> You cannot take the real exam on mobile, but our practice exam you can give on mobile.<\/li>\n\n\n\n<li><strong>Calculator<\/strong>: Below the Test, you will see a Desmos calculator and graph for Math. The same, Desmos, will be used in real exams, so learn &#8220;How to use Desmos Calculator.&#8221;<\/li>\n\n\n\n<li><strong>Tips:<\/strong> This article will help you learn more about the SAT Exams. <a href=\"https:\/\/us.mrenglishkj.com\/sat\/everything-about-the-sat\/\" target=\"_blank\" rel=\"noopener\" title=\"SAT: EVERYTHING ABOUT THE SAT\">SAT: EVERYTHING ABOUT THE SAT<\/a><\/li>\n<\/ol>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<figure class=\"wp-block-embed alignfull is-type-wp-embed is-provider-genially wp-block-embed-genially\"><div class=\"wp-block-embed__wrapper\">\n<iframe class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" title=\"SAT 6 Math Module 1st\" frameborder='0' width='1200' height='675' src='https:\/\/view.genially.com\/6753390374eab89fb913f7cc#?secret=RXXxsXM4nr' data-secret='RXXxsXM4nr' scrolling='yes'><\/iframe>\n<\/div><figcaption class=\"wp-element-caption\">Wait here for the SAT Test to appear.<\/figcaption><\/figure>\n\n\n\n<div style=\"height:40px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<p>Our team has reviewed some of the best SAT learning materials for your convenience. These materials are best for your career growth.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Check Our Review Blog: <a href=\"https:\/\/review.mrenglishkj.com\/\" target=\"_blank\" rel=\"noopener\" title=\"\">review.mrenglishkj.com<\/a><\/li>\n<\/ul>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<!-- HTML for the Desmos Calculator Embed (Always Visible) -->\n<div id=\"desmos-container\">\n    <iframe loading=\"lazy\"\n        src=\"https:\/\/www.desmos.com\/calculator\/fxgemyy2gl\"\n        width=\"100%\"\n        height=\"500px\"\n        frameborder=\"0\"\n        allowfullscreen\n    ><\/iframe>\n<\/div>\n\n<!-- Button to Open Calculator in Slide-Out Panel -->\n<button id=\"desmos-toggle\" style=\"position: fixed; top: 20px; right: 20px; z-index: 1000;\">\n    Open Calculator\n<\/button>\n\n<!-- Slide-Out Desmos Calculator Panel (hidden initially) -->\n<div id=\"desmos-panel\">\n    <iframe loading=\"lazy\"\n        src=\"https:\/\/www.desmos.com\/calculator\/fxgemyy2gl\"\n        width=\"100%\"\n        height=\"95%\"\n        frameborder=\"0\"\n        allowfullscreen\n    ><\/iframe>\n<\/div>\n\n<!-- CSS Styling for the Slide-Out Panel -->\n<style>\n    \/* Main Container Styling *\/\n    #desmos-container {\n        max-width: 600px; \/* Adjust as needed *\/\n        margin: 20px auto;\n    }\n\n    \/* Slide-Out Panel Styling *\/\n    #desmos-panel {\n        position: fixed;\n        top: 0;\n        right: -400px; \/* Hidden by default *\/\n        width: 400px; \/* Adjust width as needed *\/\n        height: 100vh;\n        background-color: white;\n        border-left: 1px solid #ccc;\n        box-shadow: -2px 0 5px rgba(0, 0, 0, 0.2);\n        transition: right 0.3s ease;\n        z-index: 999; \/* Ensure it overlays content *\/\n    }\n\n    #desmos-panel.open {\n        right: 0;\n    }\n<\/style>\n\n<!-- JavaScript to Toggle the Slide-Out Panel -->\n<script>\n    document.getElementById(\"desmos-toggle\").onclick = function() {\n        var panel = document.getElementById(\"desmos-panel\");\n        if (panel.classList.contains(\"open\")) {\n            panel.classList.remove(\"open\");\n        } else {\n            panel.classList.add(\"open\");\n        }\n    };\n<\/script>\n\n\n\n<p class=\"has-text-align-center has-small-font-size\">Wait for the Desmos Calculator to appear.<\/p>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<h3 class=\"wp-block-heading\">SAT MATH TEST ANSWERS WITH EXPLANATION AND TIPS<\/h3>\n\n\n\n<p>Do not open the tabs before finishing the practice test above! For your convenience, we have compiled all the solutions and their explanations here. We will also give you some tips or advice to help you understand them better. You&#8217;ll see <strong>&#8216;why this answer is correct&#8217;<\/strong> and <strong>&#8216;why this is incorrect.&#8217;<\/strong><\/p>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<h4 class=\"wp-block-heading\">Math Solutions and Explanations:<\/h4>\n\n\n\n<p>The light red color shows the Question, green shows the Correct answer, red shows the Incorrect one, and blue shows Tips or Tricks with step-by-step explanations.<\/p>\n\n\n\n<div class=\"wp-block-coblocks-accordion alignfull\">\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>1st Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> What is 10% of 470?<br>A) 37<br>B) 47<br>C) 423<br>D) 460<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice B<\/strong> is correct. 10% of a quantity means 10\/100 times the quantity. Therefore, 10% of 470 can be represented as 10\/100 (470), which is equivalent to 0.10(470), or 47. Therefore, 10% of 470 is 47.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect. This is 10% of 370, not 10% of 470.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect. This is 90% of 470, not 10% of 470.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect. This is 470 &#8211; 10, not 10% of 470.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">It is a simple percentage calculation, it is best to practice using the Desmos calculator.<br><br>470 divided by 10%<br>470 x 10\/100<br>4700\/100<br><strong>47 Answer.<br><\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>2nd Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>4<em>x<\/em> + 6 = 18<br>Question:<\/strong> Which equation has the same solution as the given equation?<br>A) 4<em>x<\/em> = 108<br>B) 4<em>x<\/em> = 24<br>C) 4<em>x<\/em> = 12<br>D) 4<em>x<\/em> = 3<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice C<\/strong> is correct. Subtracting 6 from both sides of the given equation yields 4x =12, which is the equation given in choice C. Since this equation is equivalent to the given equation, it has the same solution as the given equation.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">It is also a simple one.<br><strong>4<em>x<\/em> + 6 = 18<\/strong> (When one thing moves from one side to another, it changes its sign from &#8216;negative to positive&#8217; or &#8216;positive to negative.&#8217;)<strong><br><\/strong>4<em>x<\/em> = 18 &#8211; 6<br><strong>4<em>x<\/em> = 12<\/strong> (Option C: We found our answer but what if we were asked to find the value of <em>x<\/em>.)<br>Then, 4<em>x<\/em> = 12<br><em>x<\/em> = 12\/4<br><em>x<\/em> = 3.<br><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>3rd Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> If <em>y<\/em> = 5<em>x<\/em> + 10, what is the value of <em>y<\/em> when <em>x<\/em> = 8?<br>A) 23<br>B) -23<br>C) -50<br>D) 50<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice D<\/strong> is correct. The correct answer is 50. Substituting 8 for <em>x<\/em> in the given equation yields <em>y<\/em> = 5(8) + 10, or <em>y<\/em> = 50. Therefore, the value of <em>y<\/em> is 50 when <em>x<\/em> = 8.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><em>y<\/em> = 5<em>x<\/em> + 10<br>Put the value of <em>x<\/em> which is 8. (<em>x<\/em> = 8)<br><em>y<\/em> = 5(8) + 10<br><em>y<\/em> = 40 + 10<br><strong><em>y<\/em> = 50.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>4th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"293\" height=\"187\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-35.png\" alt=\"Learn and practice Graph for the SAT\" class=\"wp-image-4925\"\/><\/figure>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> The bar graph shows the distribution of 419 cans collected by 10 different groups for a food drive. How many cans were collected by group 6?<br>A) 40<br>B) 267<br>C) 266<br>D) 6<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Option A <\/strong>is<strong> <\/strong>the correct answer. The correct answer is 40. The height of each bar in the bar graph shown represents the number of cans collected by the group specified at the bottom of the bar. The bar for group 6 reaches a height of 40. Therefore, group 6 collected 40 cans.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">The question only asks for Group 6, not 1 to 6 or so. So the bar graph shows, a horizontal line, and there is the number 6 (Group 6). On vertically, you will see the bar goes to the line of 40 (Number of Cans).<\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>5th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> The table gives the distribution of votes for a new school mascot and grade level for 80 students.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"305\" height=\"379\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-37.png\" alt=\"SAT question and answer take test and learn from solutions\" class=\"wp-image-4930\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-37.png 305w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-37-241x300.png 241w\" sizes=\"auto, (max-width: 305px) 100vw, 305px\" \/><\/figure>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice C <\/strong>is correct. If one of these students is selected at random, the probability of selecting a student whose vote for the new mascot was for a lion is given by the number of votes for a lion divided by the total number of votes. The given table indicates that the number of votes for a lion is 20 votes, and the total number of votes is 80 votes. The table gives the distribution of votes for 80 students, and the table shows a total of 80 votes were counted. It follows that each of the 80 students voted exactly once. Thus, the probability of selecting a student whose vote for the new mascot was for a lion is 20\/80, or 1\/4<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">In order to answer correctly, we understand the question and the table. The question asks to find the probability of getting the lion out of the total number of students.<br>We know from the table the total number of students is: 80<br>The students who vote for the lion are: 9 + 2 + 9 = 20<br><br>The formula of probability is:<br><br> <strong> Number of favorable outcomes<\/strong>             20<br><strong><sup>______________________________________________<\/sup><\/strong> =   <sup>_______<\/sup><br><strong>Total number of possible outcomes<\/strong>        80<br><br>= 20\/80<br>= 1\/4.<br><br><strong>Final Answer: C) 1\/4.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>6th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"261\" height=\"251\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-38.png\" alt=\"\" class=\"wp-image-4935\"\/><\/figure>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> The graph represents the total charge, in dollars, by an electrician for <em>x<\/em> hours of work. The electrician charges a one-time fee plus an hourly rate. What is the best interpretation of the slope of the graph?<br>A) The electrician\u2019s hourly rate<br>B) The electrician\u2019s one-time fee<br>C) The maximum amount that the electrician charges<br>D) The total amount that the electrician charges<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice A<\/strong> is correct. It\u2019s given that the electrician charges a one-time fee plus an hourly rate. It\u2019s also given that the graph represents the total charge, in dollars, for <em>x<\/em> hours of work. This graph shows a linear relationship in the <em>xy<\/em>-plane. Thus, the total charge <em>y<\/em>, in dollars, for <em>x<\/em> hours of work can be represented as <em>y<\/em> = <em>mx<\/em> + <em>b<\/em>, where <em>m<\/em> is the slope and (0, <em>b<\/em>) is the <em>y<\/em>-intercept of the graph of the equation in the <em>xy<\/em>-plane. Since the given graph represents the total charge, in dollars, by an electrician for <em>x<\/em> hours of work, it follows that its slope is <em>m<\/em>, or the electrician\u2019s hourly rate.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect. The electrician\u2019s one-time fee is represented by the <em>y<\/em>-coordinate of the <em>y<\/em>-intercept, not the slope, of the graph.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">First, let&#8217;s understand the question, the graph only has a slope but no data. It is an option-based question, which means, our answer is hidden in options. We need to find what the slope of the graph represents.<br>Let&#8217;s apply the options:<br><strong>Note:<\/strong> We must understand one thing that a slope means both <em>x<\/em> and <em>y<\/em> sides are increasing.<br><br><strong>A) The electrician\u2019s hourly rate:<br><\/strong>It is the most accurate option, cause an hourly rate means both the hour and price are increasing.<br><br><strong>B) The electrician\u2019s one-time fee:<br><\/strong>A one-time fee can&#8217;t grow and become a slope.<br><br><strong>C) The maximum amount that the electrician charges:<br><\/strong>Again, it is a growing slope from both sides, not one side.<br><br><strong>D) The total amount that the electrician charges:<br><\/strong>It is also not possible. <\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>7th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> Square X has a side length of 12 centimeters. The perimeter of square Y is 2 times the perimeter of square X. What is the length, in centimeters, of one side of square Y?<br>A) 6<br>B) 10<br>C) 14<br>D) 24<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice D<\/strong> is correct. The perimeter, <em>P<\/em>, of a square can be found using the formula <em>P<\/em> = 4<em>s<\/em>, where <em>s<\/em> is the length of each side of the square. It\u2019s given that square <em>X<\/em> has a side length of 12 centimeters. Substituting 12 for <em>s<\/em> in the formula for the perimeter of a square yields <em>P<\/em> = 4(12), or <em>P<\/em> = 48. Therefore, the perimeter of square <em>X<\/em> is 48 centimeters. It\u2019s also given that the perimeter of square <em>Y<\/em> is 2 times the perimeter of square <em>X<\/em>. Therefore, the perimeter of square <em>Y<\/em> is 2(48), or 96, centimeters. Substituting 96 for <em>P<\/em> in the formula <em>P<\/em> = 4<em>s<\/em> gives 96 = 4<em>s<\/em>. Dividing both sides of this equation by 4 gives 24 = <em>s<\/em>. Therefore, the length of one side of square Y is 24 centimeters.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem:<\/strong><br>Square X has a side length of 12 cm. The perimeter of square Y is <strong>2 times the perimeter of square X<\/strong>. What is the length, in centimeters, of one side of square Y?<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Understand the relationship between the side length and perimeter of a square<\/strong><br>The perimeter of a square is given by the formula:<br>Perimeter = 4 \u00d7 side&nbsp;length.<br><br><strong>Step 2: Calculate the perimeter of square <em>X<\/em><\/strong><br>For square X, the side length is 12 cm. Using the formula:<br>Perimeter&nbsp;of&nbsp;square&nbsp;X = 4 \u00d7 12 = 48\u2009cm.<br><br><strong>Step 3: Relate the perimeter of square Y to square X<\/strong><br>We are told that the perimeter of square Y is <strong>2 times the perimeter of square X<\/strong>:<br>Perimeter&nbsp;of&nbsp;square&nbsp;Y = 2 \u00d7 48 = 96\u2009cm.<br><br><strong>Step 4: Determine the side length of the <\/strong>square Y<br>Using the perimeter formula for a square, solve for the side length of square Y:<br>Perimeter&nbsp;of&nbsp;square&nbsp;Y = 4 \u00d7 side&nbsp;length&nbsp;of&nbsp;square&nbsp;Y.<br>Substitute the known perimeter of square Y (96 cm):<br>96 = 4 \u00d7 side&nbsp;length&nbsp;of&nbsp;square&nbsp;Y.<br>Divide both sides by 4:<br>The side&nbsp;length&nbsp;of&nbsp;square&nbsp;Y = 96\/4 = 24\u2009cm.<br><br><strong>Verification:<\/strong><br>1. Calculate the perimeter of square Y using the side length of 24 cm:<br>4 \u00d7 24 = 96\u2009cm.<br>This matches the given condition that square Y\u2019s perimeter is <strong>2 times<\/strong> that of square X.<br>2. The solution is consistent with the problem conditions.<br><br><strong>Final Answer:<\/strong> The side length of square Y is <strong>24 centimeters<\/strong>.<\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>8th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> An object is kicked from a platform. The equation <em>h<\/em> represents this situation, where <strong><em>h<\/em> = -4.9<em>t<\/em><sup>2<\/sup> + 7<em>t<\/em> + 9<\/strong> is the height of the object above the ground, in meters, <em>t<\/em> seconds after it is kicked. Which number represents the height, in meters, from which the object was kicked?<br>A) 0<br>B) 4.9<br>C) 7<br>D) 9<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice D<\/strong> is correct. It\u2019s given that the equation <strong><em>h<\/em> = -4.9<em>t<\/em><sup>2<\/sup> + 7<em>t<\/em> + 9<\/strong> represents this situation, where h is the height, in meters, of the object <em>t<\/em> seconds after it is kicked. It follows that the height, in meters, from which the object was kicked is the value of h when <em>t<\/em> = 0. Substituting 0 for <em>t<\/em> in the equation <strong><em>h<\/em> = -4.9<em>t<\/em><sup>2<\/sup> + 7<em>t<\/em> + 9<\/strong> yields <em>h<\/em> = -4.9(0)<sup>2<\/sup> + 7(0) + 9, or <em>h<\/em> = 9. Therefore, the object was kicked from a height of 9 meters.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem:<\/strong><br>The height <em>h<\/em> of an object kicked from a platform is modeled by the equation:<br><em>h<\/em> = \u22124.9<em>t<\/em><sup>2<\/sup> + 7<em>t<\/em> + 9,<br>where <em>h<\/em> is the height above the ground (in meters), and <em>t<\/em> is the time (in seconds) after the object is kicked.<br>We are asked to find <strong>the height from which the object was kicked<\/strong>.<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Understand the question<\/strong><br>The <strong>height from which the object was kicked<\/strong> refers to the initial height of the object above the ground at <em>t<\/em> = 0 (the moment it is kicked). This is the <strong>starting height<\/strong>, which can be found by substituting <em>t<\/em> = 0 into the equation for <em>h<\/em>.<br><br><strong>Step 2: Substitute <em>t<\/em> = 0 into the equation<\/strong><br>The given equation is:<br><em>h<\/em> = \u22124.9<em>t<\/em><sup>2<\/sup> + 7<em>t<\/em> + 9.<br>Substitute <em>t<\/em> = 0:<br><em>h<\/em> = \u22124.9(0)<sup>2<\/sup> + 7(0) + 9.<br><br><strong>Step 3: Simplify the equation<\/strong><br>Simplify each term:<br><em>h<\/em> = \u22124.9(0) + 7(0) + 9.<br><em>h<\/em> = 0 + 0 + 9.<br><em>h<\/em> = 9.<br><br><strong>Step 4: Interpret the result<\/strong><br>At <em>t<\/em> = 0, the height of the object is <strong>9 meters<\/strong>. This means the object was kicked from a platform 9 meters above the ground.<br><br><strong>Verification:<\/strong><br>1. At <em>t<\/em> = 0, only the constant term +9 in the equation contributes to the height. This is because the terms \u22124.9<em>t<\/em><sup>2<\/sup> and 7<em>t<\/em> both equal zero when <em>t<\/em> = 0.<br>2. The constant term in a quadratic height equation like this always represents the initial height (starting height).<br><br><strong>Final Answer:<\/strong> The height from which the object was kicked is <strong>9 meters<\/strong>.<\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>9th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><em>f<\/em>(<em>x<\/em>) = 4<em>x<\/em><sup>2<\/sup> \u2212 50<em>x<\/em> + 126<br><strong>Question:<\/strong> The given equation defines the function <em>f<\/em>. For what value of <em>x<\/em> does <em>f<\/em>(x) reach its minimum?<br>A) 6.25<br>B) 16<br>C) 6.5<br>D) 16.25<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>The correct answer is Option A.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"488\" height=\"262\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-39.png\" alt=\"Solutions of math problems and SAT tests\" class=\"wp-image-4949\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-39.png 488w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-39-300x161.png 300w\" sizes=\"auto, (max-width: 488px) 100vw, 488px\" \/><\/figure>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Recognize the nature of the function<\/strong><br>The given function is a <strong>quadratic function<\/strong> of the form:<br><em>f<\/em>(<em>x<\/em>) = <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em>,<br>where:<br>~ <em>a<\/em> = 4,<br>~ <em>b<\/em> = \u221250,<br>~ <em>c<\/em> = 126.<br>Since the coefficient of <em>x<\/em><sup>2<\/sup>(<em>a<\/em> = 4) is <strong>positive<\/strong>, the parabola opens upwards, meaning the vertex represents the <strong>minimum value<\/strong> of the function. <\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"615\" height=\"594\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-40.png\" alt=\"Solve sat math problems in details\" class=\"wp-image-4959\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-40.png 615w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-40-300x290.png 300w\" sizes=\"auto, (max-width: 615px) 100vw, 615px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Verification of the Solution<\/strong><br>To verify, substitute <em>x<\/em> = 6.25 into the equation <em>f<\/em>(<em>x<\/em>) to confirm that it represents the minimum:<br>1. Compute <em>f<\/em>(6.25):<br><em>f<\/em>(6.25) = 4(6.25)<sup>2<\/sup> \u2212 50(6.25) + 126.<br>2. Simplify (6.25)<sup>2<\/sup>:<br>(6.25)<sup>2<\/sup> = 39.0625.<br>3. Substitute back:<br><em>f<\/em>(6.25) = 4(39.0625) \u2212 50(6.25) + 126.<br><em>f<\/em>(6.25) = 156.25 \u2212 312.5 + 126.<br><em>f<\/em>(6.25) = \u221230.25.<br>Thus, the minimum value of the function is \u221230.25 at <em>x<\/em> = 6.25. This confirms our calculation.<br><br><strong>Final Answer: A) 6.25.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>10th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> Vivian bought party hats and cupcakes for $71. Each package of party hats costs $3, and each cupcake costs $1. If Vivian bought 10 packages of party hats, how many cupcakes did she buy?<br>A) 4<br>B) 96<br>C) 20<br>D) 41<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice D<\/strong> is correct. The correct answer is 41. The number of cupcakes Vivian bought can be found by first finding the amount Vivian spent on cupcakes. The amount Vivian spent on cupcakes can be found by subtracting the amount Vivian spent on party hats from the total amount Vivian spent. The amount Vivian spent on party hats can be found by multiplying the cost per package of party hats by the number of packages of party hats, which yields $3 <strong><sup><sub>.<\/sub><\/sup><\/strong> 10, or $30. Subtracting the amount Vivian spent on party hats, $30, from the total amount Vivian spent, $71, yields $71 &#8211; $30, or $41. Since the amount Vivian spent on cupcakes was $41 and each cupcake cost $1, it follows that Vivian bought 41 cupcakes.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">We are tasked with determining how many cupcakes Vivian bought. Let&#8217;s solve this step by step and verify the solution.<br><br><strong>Step-by-Step Solution<\/strong><br><strong>Step 1: Define the variables<\/strong><br>~ Let the number of cupcakes Vivian bought be <em>x<\/em>.<br>~ The cost of each package of party hats is 3.<br>~ The cost of each cupcake is 1.<br><br><strong>Step 2: Write an equation for the total cost<\/strong><br>Vivian spent $71 total, and she bought 10 packages of party hats. Therefore, the total cost can be expressed as:<br>3(10) + 1(<em>x<\/em>) = 71.<br><br><strong>Step 3: Simplify the equation<\/strong><br>Simplify the terms:<br>30 + <em>x<\/em> = 71.<br><br><strong>Step 4: Solve for <em>x<\/em><\/strong><br>Subtract 30 from both sides:<br><em>x<\/em> = 71 \u2212 30.<br><em>x<\/em> = 41.<br>Thus, Vivian bought <strong>41 cupcakes<\/strong>.<br><br><strong>Verification<\/strong><br>To verify, calculate the total cost:<br>1. The cost of 10 packages of party hats:<br>10 \u00d7 3 = 30.<br>2. The cost of 41 cupcakes:<br>41 \u00d7 1 = 41.<br>3. Total cost:<br>30 + 41 = 71.<br>This matches the given total cost of $71. The solution is correct.<br><br><strong>Final Answer: D) 41.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>11th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>z<sup>2<\/sup> + 10<em>z<\/em> &#8211; 24 = 0<\/strong><br><strong>Question:<\/strong> What is one of the solutions to the given equation?<br>A) 2 or -12<br>B) 4 or -24<br>C) -2 or 12<br>D) -4 or 24<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice A<\/strong> is correct. The correct answer is either 2 or -12. The left-hand side of the given equation can be rewritten by factoring. The two values that multiply to -24 and add to 10 are 12 and -2. It follows that the given equation can be rewritten as (<em>z<\/em> + 12)(<em>z<\/em> &#8211; 2) = 0. Setting each factor equal to 0 yields two equations: <em>z<\/em> +12 = 0 and <em>z<\/em> &#8211; 2 = 0. Subtracting 12 from both sides of the equation <em>z<\/em> + 12 = 0 results in <em>z<\/em> = -12. Adding 2 to both sides of the equation <em>z<\/em> &#8211; 2 = 0 results in <em>z<\/em> = 2. Note that 2 and -12 are examples of ways to enter a correct answer.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">There are two ways to do this equation.<br>1. Quadratic Method<br>2. Factorize Method<br>I will show you both methods and tell you when can you use them.<br>To identify, an equation should look like this: <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0<br>Notice, the question looks the same: <strong>z<sup>2<\/sup> + 10<em>z<\/em> &#8211; 24 = 0<\/strong><br><br><strong>Let&#8217;s use the Quadratic Method:<br>Step-by-Step Solution:<\/strong><br><strong>Step 1: Identify the quadratic equation structure.<\/strong><br>The given equation is in the standard form:<br><em>az<\/em><sup>2<\/sup> + <em>bz<\/em> + <em>c<\/em> = 0<br>Here:<br>~ <em>a<\/em> = 1 (coefficient of <em>z<\/em><sup>2<\/sup>),<br>~ <em>b<\/em> = 10 (coefficient of zzz),<br>~ <em>c<\/em> = \u221224 (constant term).<br><br><strong>Step 2: Solve using the quadratic formula.<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"536\" height=\"659\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-43.png\" alt=\"Quadratic formula explanation using sat math example\" class=\"wp-image-4981\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-43.png 536w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-43-244x300.png 244w\" sizes=\"auto, (max-width: 536px) 100vw, 536px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">( <span style=\"text-decoration: underline\"><strong>+<\/strong><\/span> ) This sign means that it can be positive or negative, so let&#8217;s try both ways.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"678\" height=\"649\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-44.png\" alt=\"Quadratic method solutions using sat math test\" class=\"wp-image-4982\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-44.png 678w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-44-300x287.png 300w\" sizes=\"auto, (max-width: 678px) 100vw, 678px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Final Answer: A) 2 or -12.<\/strong><br><br><strong>Let&#8217;s use the Factorize Method:<br>Step-by-Step Solution:<\/strong><br><strong>Step 1: Identify the equation structure (<em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0).<\/strong><br>The given quadratic equation is:<br><em>z<\/em><sup>2<\/sup> + 10<em>z<\/em> \u2212 24 = 0<br>Here, the goal is to factorize it into the form:<br>(<em>z<\/em> + <em>p<\/em>)(<em>z<\/em> + <em>q<\/em>) = 0<br>where <em>p<\/em> and <em>q<\/em> are numbers such that:<br>1. Their product is <em>c<\/em> = \u221224.<br>2. Their sum is <em>b<\/em> = 10.<br><br><strong>Step 2: Find two numbers that satisfy the conditions.<\/strong><br>We need two numbers whose:<br>~ Product = \u221224,<br>~ Sum = 10.<br>By trial or inspection: <br><em>p<\/em> = 12 and <em>q<\/em> = \u22122<br>because:<br>12 <strong>\u22c5<\/strong> (\u22122) = \u221224 (<em>c<\/em> comes by multiplying two numbers)<br>12 + (\u22122) = 10 (<em>b<\/em> comes by adding the same two numbers)<br>It must be the same numbers you multiply or add and it should give <em>c<\/em> and <em>b<\/em>. If it doesn&#8217;t happen then it is not a factor and you cannot solve the equation using this method.<br><br><strong>Step 3: Write the equation in factored form.<\/strong><br>Using <em>p<\/em> = 12 and <em>q<\/em> = \u22122, the quadratic equation becomes:<br><em>z<\/em><sup>2<\/sup> + 10<em>z<\/em> \u2212 24 = (<em>z<\/em> + 12)(<em>z<\/em> \u2212 2) = 0<br><br><strong>Step 4: Solve for <em>z<\/em>.<\/strong><br>Set each factor equal to 0:<br>1. <em>z <\/em>+ 12 = 0<br><em>z<\/em> = \u221212<br>2. <em>z<\/em> \u2212 2 = 0<br><em>z<\/em> = 2<br>Thus, the two solutions are:<br><em>z<\/em> = \u221212 and <em>z<\/em> = 2<br><br><strong>Step 5: Verify the solutions.<\/strong><br>Substitute both <em>z<\/em> = 2 and <em>z<\/em> = \u221212 back into the original equation <em>z<\/em><sup>2<\/sup> + 10<em>z<\/em> \u2212 24 = 0:<br>1. For <em>z<\/em> = 2:<br>(2)<sup>2<\/sup> + 10(2) \u2212 24 = 4 + 20 \u2212 24 = 0<br>2. For <em>z<\/em> = \u221212:<br>(\u221212)<sup>2<\/sup> + 10(\u221212) \u2212 24 = 144 \u2212 120 \u2212 24 = 0<br>Both solutions are correct.<br><br><strong>Final Answer:<\/strong> The two solutions are <strong><em>z<\/em> = 2 and <em>z<\/em> = \u221212<\/strong>.<\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>12th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> Bacteria are growing in a liquid growth medium. There were 300,000 cells per milliliter during an initial observation. The number of cells per milliliter doubles every 3 hours. How many cells per milliliter will there be 15 hours after the initial observation?<br>A) 1,500,000<br>B) 2,400,000<br>C) 4,500,000<br>D) 9,600,000<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice D<\/strong> is correct. <\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"494\" height=\"313\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-42.png\" alt=\"Prepare for the SAT test for free and take SATs\" class=\"wp-image-4970\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-42.png 494w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-42-300x190.png 300w\" sizes=\"auto, (max-width: 494px) 100vw, 494px\" \/><\/figure>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem Breakdown:<\/strong><br>We are tasked with calculating the number of bacteria cells per milliliter after 15 hours, given:<br>1. Initially, there are 300,000 cells per milliliter.<br>2. The number of cells doubles every 3 hours.<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Understand the growth pattern<\/strong><br>The problem states that the number of cells doubles every 3 hours. This is an example of exponential growth, where the amount of growth depends on a doubling factor for a given time.<br>The formula for this type of growth is:<br><em>N<\/em>(<em>t<\/em>) = N<sub>0<\/sub> <strong>\u22c5<\/strong> 2<em><sup>t\/T<\/sup><\/em>,<br>where:<br>~ <em>N<\/em>(<em>t<\/em>) = the number of cells after <em>t<\/em> hours,<br>~ <em>N<\/em><sub>0<\/sub>\u200b = the initial number of cells (300,000),<br>~ <em>T<\/em> = the doubling time (3 hours),<br>~ <em>t<\/em> = total time elapsed (15 hours in this case).<br><br><strong>Step 2: Substitute the known values into the formula<\/strong><br>We are solving for <em>N<\/em>(15):<br><em>N<\/em>(15) = 300,000 <strong>\u22c5<\/strong> 2<em><sup>15\/3<\/sup><\/em>.<br>Here:<br>~ N<sub>0<\/sub> = 300,000,<br>~ <em>T<\/em> = 3,<br>~ <em>t<\/em> = 15.<br><br><strong>Step 3: Simplify the exponent<\/strong><br>The exponent is:<br>15\/3 = 5.<br>Thus:<br><em>N<\/em>(15) = 300,000 <strong>\u22c5<\/strong> 2<sup>5<\/sup>.<br><br><strong>Step 4: Calculate 2<sup>5<\/sup><\/strong><br>2<sup>5<\/sup> = 2 \u22c5 2 \u22c5 2 \u22c5 2 \u22c5 2 = 32.<br>Substituting this value:<br><em>N<\/em>(15) = 300,000 <strong>\u22c5<\/strong> 32.<br><br><strong>Step 5: Perform the multiplication<\/strong><br>300,000 <strong>\u22c5<\/strong> 32 = 9,600,000.<br><br><strong>Verification:<\/strong><br>Let\u2019s confirm step-by-step:<br>1. Doubling every 3 hours means the cells will double 15\/3 = 5 times.<br>2. The initial count is 300,000. Doubling it 5 times results in:<br>~ After 3 hours: 300,000 <strong>\u22c5<\/strong> 2 = 600,000,<br>~ After 6 hours: 600,000 <strong>\u22c5<\/strong> 2 = 1,200,000,<br>~ After 9 hours: 1,200,000 <strong>\u22c5<\/strong> 2 = 2,400,000,<br>~ After 12 hours: 2,400,000 \u22c5 2 = 4,800,000,<br>~ After 15 hours: 4,800,000 <strong>\u22c5<\/strong> 2 = 9,600,000,<br>The calculation matches the formula-based answer.<br><br><strong>Final Answer:<\/strong><br>After 15 hours, the number of cells per milliliter will be: <strong>9,600,000.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>13th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> The product of two positive integers is 546. If the first integer is 11 greater than twice the second integer, what is the smaller of the two integers?<br>A) 21<br>B) 28<br>C) 32<br>D) 35<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice A<\/strong> is correct. The <em>x<\/em>-intercept of a graph in the <em>xy<\/em>-plane is the point on the graph where <em>y<\/em> = 0. It\u2019s given that function <em>h<\/em> is defined by <em>h<\/em>(<em>x<\/em>) = 4<em>x<\/em> + 28. Therefore, the equation representing the graph of <em>y<\/em> = <em>h<\/em>(<em>x<\/em>) is <em>y<\/em> = 4<em>x<\/em> + 28. Substituting 0 for <em>y<\/em> in the equation <em>y<\/em> = 4<em>x<\/em> + 28 yields 0 = 4<em>x<\/em> + 28. Subtracting 28 from both sides of this equation yields -28 = 4<em>x<\/em>. Dividing both sides of this equation by 4 yields -7 = x. Therefore, the <em>x<\/em>-intercept of the graph of <em>y<\/em> = <em>h<\/em>(<em>x<\/em>) in the <em>xy<\/em>-plane is (-7, 0). It\u2019s given that the <em>x<\/em>-intercept of the graph of <em>y<\/em> = <em>h<\/em>(<em>x<\/em>) is (<em>a<\/em>, 0). Therefore, <em>a<\/em> = -7. The <em>y<\/em>-intercept of a graph in the <em>xy<\/em>-plane is the point on the graph where <em>x<\/em> = 0. Substituting 0 for <em>x<\/em> in the equation <em>y<\/em> = 4<em>x<\/em> + 28 yields <em>y<\/em> = 4(0) + 28, or y = 28. Therefore, the <em>y<\/em>-intercept of the graph of <em>y<\/em> = <em>h<\/em>(<em>x<\/em>) in the <em>xy<\/em>-plane is (0, 28). It\u2019s given that the <em>y<\/em>-intercept of the graph of <em>y<\/em> = <em>h<\/em>(<em>x<\/em>) is (0, <em>b<\/em>). Therefore, <em>b<\/em> = 28. If <em>a<\/em> = -7 and <em>b<\/em> = 28, then the value of <em>a<\/em> + <em>b<\/em> is -7 + 28, or 21.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect. This is the value of <em>b<\/em>, not <em>a<\/em> + <em>b<\/em>.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect. This is the value of &#8211;<em>a<\/em> + <em>b<\/em>, not <em>a<\/em> + <em>b<\/em>.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem Breakdown:<\/strong><br>We are tasked with finding the value of <em>a<\/em> + <em>b<\/em>, where:<br>~ <em>a<\/em> is the <em>x<\/em>-intercept of the graph of <em>y<\/em> = <em>h<\/em>(<em>x<\/em>).<br>~ <em>b<\/em> is the <em>y<\/em>-intercept of the graph of <em>y<\/em> = <em>h<\/em>(<em>x<\/em>). The function is defined as <em>h<\/em>(<em>x<\/em>) = 4<em>x<\/em> + 28.<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Find the <em>x<\/em>-intercept (<em>a<\/em>)<\/strong><br>The <em>x<\/em>-intercept occurs when <em>y<\/em> = 0. Substituting <em>h<\/em>(<em>x<\/em>) = 0 into the function:<br>0 = 4<em>x<\/em> + 28.<br>Solve for <em>x<\/em>:<br>4<em>x<\/em> = \u221228,<br><em>x<\/em> = \u22127.<br>Thus, the <em>x<\/em>-intercept is <em>a<\/em> = \u22127.<br><br><strong>Step 2: Find the <em>y<\/em>-intercept (<em>b<\/em>)<\/strong><br>The <em>y<\/em>-intercept occurs when <em>x<\/em> = 0. Substituting <em>x<\/em> = 0 into the function:<br><em>h<\/em>(0) = 4(0) + 28.<br>Simplify: <em>h<\/em>(0) = 28.<br>Thus, the <em>y<\/em>-intercept is <em>b<\/em> = 28.<br><br><strong>Step 3: Find <em>a<\/em> + <em>b<\/em><\/strong><br>Substitute the values of <em>a<\/em> and <em>b<\/em> into <em>a<\/em> + <em>b<\/em>:<br><em>a<\/em> + <em>b<\/em> = \u22127 + 28.<br>Simplify: <em>a<\/em> + <em>b<\/em> = 21.<br><br><strong>Verification:<\/strong><br>~ For <em>x<\/em>-intercept:<br>Substituting <em>y<\/em> = 0 in 4<em>x<\/em> + 28 = 0, we found <em>x<\/em> = \u22127.<br>~ For <em>y<\/em>-intercept:<br>Substituting <em>x<\/em> = 0 in 4<em>x<\/em> + 28, we found <em>y<\/em> = 28.<br>~ <em>a<\/em> + <em>b<\/em> = \u22127 + 28 = 21.<br>Both intercepts and the result are consistent.<br><br><strong>Final Answer:<\/strong> The value of <em>a<\/em> + <em>b<\/em> is: 21.<\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>14th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> One of the factors of 2<em>x<\/em><sup>3<\/sup> + 42<em>x<\/em><sup>2<\/sup> + 208<em>x<\/em> is <em>x<\/em> + <em>b<\/em>, where <em>b<\/em> is a positive constant. What is the smallest possible value of <em>b<\/em>?<br>A) 8<br>B) 13<br>C) -8<br>D) -13<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice A<\/strong> is correct. The correct answer is 8. Since each term of the given expression, 2<em>x<\/em><sup>3<\/sup> + 42<em>x<\/em><sup>2<\/sup> + 208<em>x<\/em>, has a factor of 2<em>x<\/em>, the expression can be rewritten as 2<em>x<\/em>(<em>x<\/em><sup>2<\/sup>) + 2<em>x<\/em>(21<em>x<\/em>) + 2<em>x<\/em>(104), or 2<em>x<\/em>(<em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104). Since the values 8 and 13 have a sum of 21 and a product of 104, the expression <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104 can be factored as (<em>x<\/em> + 8)(<em>x<\/em> + 13). Therefore, the given expression can be factored as 2<em>x<\/em>(<em>x<\/em> + 8)(<em>x<\/em> + 13). It follows that the factors of the given expression are 2, <em>x<\/em>, <em>x<\/em> + 8, and <em>x<\/em> + 13. Of these factors, only <em>x<\/em> + 8 and <em>x<\/em> + 13 are of the form <em>x<\/em> + <em>b<\/em>, where <em>b<\/em> is a positive constant. Therefore, the possible values of <em>b<\/em> are 8 and 13. Thus, the smallest possible value of <em>b<\/em> is 8.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">There are two methods to solve this equation.<br>1. Quadratic Formula<br>2. Factoring Formula<br><br>1. Let&#8217;s use the <strong>Quadratic Formula:<\/strong><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Factor out the greatest common factor (GCF)<\/strong><br>Start by factoring out the GCF of the terms in the polynomial:<br>2<em>x<\/em><sup>3<\/sup> + 42<em>x<\/em><sup>2<\/sup> + 208<em>x<\/em> = 2<em>x<\/em>(<em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104).<br>Thus, the polynomial simplifies to:<br>2<em>x<\/em>(<em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104).<br>Now, we focus on the quadratic <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104 to find the factor <em>x<\/em> + <em>b<\/em>.<br><br><strong>Step 2: Use the Factor Theorem<\/strong><br>The Factor Theorem states that if <em>x<\/em> + <em>b<\/em> is a factor of <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104, then substituting <em>x<\/em> = \u2212<em>b<\/em> into the polynomial should yield 0:<br><em>f<\/em>(<em>x<\/em>) = <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104, and <em>f<\/em>(\u2212<em>b<\/em>) = 0.<br>Substitute <em>x<\/em> = \u2212<em>b<\/em> into <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104:<br>(\u2212<em>b<\/em>)<sup>2 <\/sup>+ 21(\u2212<em>b<\/em>) + 104 = 0.<br>Simplify: <em>b<\/em><sup>2<\/sup> \u2212 21<em>b<\/em> + 104 = 0.<br><br><strong>Step 3: Solve for <em>b<\/em><\/strong><br>To solve the quadratic <em>b<\/em><sup>2<\/sup> \u2212 21<em>b<\/em> + 104 = 0, we use the quadratic formula:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"530\" height=\"645\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-45.png\" alt=\"Simple SAT math problem answers and learning\" class=\"wp-image-5022\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-45.png 530w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-45-247x300.png 247w\" sizes=\"auto, (max-width: 530px) 100vw, 530px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">Thus, the two possible values of <em>b<\/em> are 13 and 8.<br><br><strong>Step 4: Identify the smallest positive <em>b<\/em><\/strong><br>The smallest positive value of <em>b<\/em> is:<br><em>b<\/em> = 8.<br><br><strong>Verification:<\/strong><br>To verify, substitute <em>x<\/em> = \u22128 into <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104 to confirm that <em>x<\/em> + 8 is indeed a factor:<br><em>f<\/em>(\u22128) = (\u22128)<sup>2<\/sup> + 21(\u22128) + 104.<br><em>f<\/em>(\u22128) = 64 \u2212 168 + 104 = 0.<br>This confirms that <em>x<\/em> + 8 is a factor of the polynomial.<br><br><strong>Final Answer:<\/strong> The smallest possible value of <em>b<\/em> is: <strong>8.<\/strong><br><br>2. Let&#8217;s use the <strong>Factoring Formula:<\/strong><br><strong>Step 1: Factor out the greatest common factor (GCF)<\/strong><br>The GCF of 2<em>x<\/em><sup>3<\/sup>, 42<em>x<\/em><sup>2<\/sup>, and 208<em>x<\/em> is 2<em>x<\/em>.<br>2<em>x<\/em><sup>3<\/sup> + 42<em>x<\/em><sup>2<\/sup> + 208<em>x<\/em> = 2<em>x<\/em>(<em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104)<br><br><strong>Step 2: Factor the quadratic <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104<\/strong><br>Now, we focus on factoring <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104.<br>We need two numbers that multiply to 104 (the constant term) and add to 21 (the coefficient of <em>x<\/em>):<br>~ The two numbers are 13 and 8, because 13 \u00d7 8 = 104 and 13 + 8 = 21.<br>So, we can factor <em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104 as:<br><em>x<\/em><sup>2<\/sup> + 21<em>x<\/em> + 104 = (<em>x<\/em> + 13)(<em>x<\/em> + 8)<br><br><strong>Step 3: Combine the factors<\/strong><br>The fully factored form of the expression is:<br>2<em>x<\/em><sup>3<\/sup> + 42<em>x<\/em><sup>2<\/sup> + 208<em>x<\/em> = 2<em>x<\/em>(<em>x<\/em> + 13)(<em>x<\/em> + 8)<br><br><strong>Step 4: Determine the smallest possible value of <em>b<\/em><\/strong><br>The factors include <em>x<\/em> + 13 and <em>x<\/em> + 8. Here, <em>b<\/em> represents the positive constant from <em>x<\/em> + <em>b<\/em>.<br>From the factors, <em>b<\/em> = 8 is the smallest possible value.<br><br><strong>Verification<\/strong><br>To verify, substitute <em>b<\/em> = 8 into <em>x<\/em> + <em>b<\/em>:<br><em>x<\/em> + 8<br>This is indeed a factor of the original expression, as shown by the factorization.<br><br><strong>Final Answer:<\/strong> The smallest possible value of <em>b<\/em> is: <strong>8.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>15th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong><em>y<\/em> = \u22121.5<br><em>y<\/em> = <em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em><\/strong><br><strong>Question:<\/strong> In the given system of equations, <em>a<\/em> is a positive constant. The system has exactly one distinct real solution. What is the value of <em>a<\/em>?<br>A) 29<br>B) 14.5<br>C) 58.5<br>D) -29<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice B<\/strong> is the correct option. The correct answer is 29\/2. According to the first equation in the given system, the value of <em>y<\/em> is -1.5. Substituting -1.5 for <em>y<\/em> in the second equation in the given system yields -1.5 = <em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em>. Adding 1.5 to both sides of this equation yields 0 = <em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em> + 1.5. If the given system has exactly one distinct real solution, it follows that 0 = <em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em> + 1.5 has exactly one distinct real solution. A quadratic equation in the form 0 = <em>px<\/em><sup>2<\/sup> + <em>qx<\/em> + <em>r<\/em>, where <em>p<\/em>, <em>q<\/em>, and <em>r<\/em> are constants, has exactly one distinct real solution if and only if the discriminant, <em>q<\/em><sup>2<\/sup> &#8211; 4<em>pr<\/em>, is equal to 0. The equation 0 = <em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em> + 1.5 is in this form, where <em>p<\/em> = 1, <em>q<\/em> = 8, and <em>r<\/em> = <em>a<\/em> + 1.5. Therefore, the discriminant of the equation 0 = <em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em> + 1.5 is (8)<sup>2<\/sup> &#8211; 4(1)(<em>a<\/em> + 1.5), or 58 &#8211; 4<em>a<\/em>. Setting the discriminant equal to 0 to solve for <em>a<\/em> yields 58 &#8211; 4<em>a<\/em> = 0. Adding 4<em>a<\/em> to both sides of this equation yields 58 = 4<em>a<\/em>. Dividing both sides of this equation by 4 yields 58\/4 = <em>a<\/em>, or 29\/2 = <em>a<\/em>. Therefore, if the given system of equations has exactly one distinct real solution, the value of <em>a<\/em> is 29\/2. Note that 29\/2 and 14.5 are examples of ways to enter a correct answer.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Step 1: Set the equations equal to each other<\/strong><br>Since both equations equal <em>y<\/em>, we set them equal to each other:<br>\u22121.5 = <em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em><br>Rearrange to form a standard quadratic equation:<br><em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + <em>a<\/em> + 1.5 = 0<br><em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + (<em>a<\/em> + 1.5) = 0<br><br><strong>Step 2: Conditions for one distinct real solution<\/strong><br>A quadratic equation has exactly one distinct real solution when its <strong>discriminant<\/strong> is zero. The discriminant (\u0394) of a quadratic equation <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0 is given by:<br>\u0394 = <em>b<\/em><sup>2<\/sup> \u2212 4<em>ac<\/em><br>Here:<br><em>a<\/em> = 1 (coefficient of <em>x<\/em><sup>2<\/sup>),<br><em>b<\/em> = 8 (coefficient of <em>x<\/em>),<br><em>c<\/em> = <em>a<\/em> + 1.5 (constant term).<br>Substitute into the discriminant formula:<br>\u0394 = 8<sup>2<\/sup> \u2212 4(1)(<em>a<\/em> + 1.5)<br>\u0394 = 64 \u2212 4(<em>a<\/em> + 1.5)<br><br><strong>Step 3: Set the discriminant equal to zero<\/strong><br>For the quadratic to have exactly one distinct real solution:<br>64 \u2212 4(<em>a<\/em> + 1.5) = 0<br>Simplify: 64 \u2212 4<em>a<\/em> \u2212 6 = 0<br>58 = 4<em>a<\/em><br><em>a<\/em> = 58\/4 <br><em>a<\/em> = 14.5<br><br><strong>Step 4: Verify the solution<\/strong><br>Substitute <em>a<\/em> = 14.5 into the quadratic equation:<br><em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + (14.5 + 1.5) = 0<br><em>x<\/em><sup>2<\/sup> + 8<em>x<\/em> + 16 = 0<br>Find the discriminant:<br>\u0394 = 8<sup>2<\/sup> \u2212 4(1)(16)<br>\u0394 = 64 \u2212 64<br>\u0394 = 0<br>Since the discriminant is zero, the equation has exactly one distinct real solution. This confirms <em>a<\/em> = 14.5 is correct.<br><br><strong>Final Answer: B) 14.5.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>16th Question<\/strong> <\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><em>f<\/em>(<em>x<\/em>) = (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> \u2212 4) <br><strong>Question:<\/strong> The function <em>f<\/em> is given. Which table of values represents <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) \u2212 3?<\/p>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Option A)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">x<\/th><th class=\"has-text-align-center\" data-align=\"center\">y<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">-6<\/td><td class=\"has-text-align-center\" data-align=\"center\">-9<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">-5<\/td><td class=\"has-text-align-center\" data-align=\"center\">-8<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Option B)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">x<\/th><th class=\"has-text-align-center\" data-align=\"center\">y<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">-6<\/td><td class=\"has-text-align-center\" data-align=\"center\">-3<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">-5<\/td><td class=\"has-text-align-center\" data-align=\"center\">-3<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">-3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Option C)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">x<\/th><th class=\"has-text-align-center\" data-align=\"center\">y<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">-6<\/td><td class=\"has-text-align-center\" data-align=\"center\">-3<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">-5<\/td><td class=\"has-text-align-center\" data-align=\"center\">-2<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Option D)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">x<\/th><th class=\"has-text-align-center\" data-align=\"center\">y<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">-6<\/td><td class=\"has-text-align-center\" data-align=\"center\">3<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">-5<\/td><td class=\"has-text-align-center\" data-align=\"center\">3<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice B<\/strong> is correct. It\u2019s given that <em>f<\/em>(<em>x<\/em>) = (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> &#8211; 4) and <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) &#8211; 3. Substituting (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> &#8211; 4) for <em>f<\/em>(<em>x<\/em>) in the equation <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) &#8211; 3 yields <em>y<\/em> = (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> &#8211; 4) &#8211; 3. Substituting -6 for <em>x<\/em> in this equation yields <em>y<\/em> = (-6 + 6)(-6 + 5)(-6 &#8211; 4) &#8211; 3, or <em>y<\/em> = &#8211; 3. Substituting -5 for <em>x<\/em> in the equation <em>y<\/em> = (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> &#8211; 4) &#8211; 3 yields <em>y<\/em> = (-5 + 6)(-5 + 5)(-5 &#8211; 4) &#8211; 3, or <em>y<\/em> = &#8211; 3. Substituting 4 for <em>x<\/em> in the equation <em>y<\/em> = (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> &#8211; 4) &#8211; 3 yields <em>y<\/em> = (4 + 6)(4 + 5)(4 &#8211; 4) &#8211; 3, or <em>y<\/em> = &#8211; 3. Therefore, when <em>x<\/em> = -6 then <em>y<\/em> = -3, when <em>x<\/em> = -5 then <em>y<\/em> = -3, and when <em>x<\/em> = 4 then <em>y<\/em> = -3. Thus, the table of values in choice B represents <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) &#8211; 3.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect. This table represents <em>y<\/em> = <em>x<\/em> &#8211; 3 rather than <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) &#8211; 3.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect. This table represents <em>y<\/em> = <em>x<\/em> + 3 rather than <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) &#8211; 3.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect. This table represents <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) + 3 rather than <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) &#8211; 3.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem:<\/strong><br>Given the function:<br><em>f<\/em>(<em>x<\/em>) = (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> \u2212 4)<br>Determine which table of values represents the function:<br><em>y<\/em> = <em>f<\/em>(<em>x<\/em>) \u2212 3<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Understand the Given Functions<\/strong><br>1. <strong>Original Function:<\/strong><br><em>f<\/em>(<em>x<\/em>) = (<em>x<\/em> + 6)(<em>x<\/em> + 5)(<em>x<\/em> \u2212 4)<br>This is a <strong>cubic function<\/strong> with roots at <em>x<\/em> = \u22126, <em>x<\/em> = \u22125, and <em>x<\/em> = 4.<br>2. <strong>Modified Function:<\/strong><br><em>y<\/em> = <em>f<\/em>(<em>x<\/em>) \u2212 3<br>This transformation <strong>shifts the graph of <em>f<\/em>(<em>x<\/em>)<\/strong> <strong>downward by 3 units<\/strong>. For every <em>x<\/em>, the new <em>y<\/em>-value is 3 less than the original <em>f<\/em>(<em>x<\/em>)-value.<br><br><strong>Step 2: Identify Key Points to Populate the Table<\/strong><br>To construct a table of values for <em>y<\/em> = <em>f<\/em>(<em>x<\/em>) \u2212 3, we need to evaluate <em>y<\/em> at specific <em>x<\/em>-values. Typically, these <em>x<\/em>-values are chosen based on the roots and other significant points of the original function.<br><br><strong>Key <em>x<\/em>-values to consider:<\/strong><br>~ <strong>Roots of <em>f<\/em>(<em>x<\/em>):<\/strong> <em>x<\/em> = \u22126, <em>x<\/em> = \u22125, <em>x<\/em> = 4<br>~ <strong>Additional Points:<\/strong> Choose other <em>x<\/em>-values if needed for completeness<br><br><strong>Step 3: Calculate <em>y<\/em> for Each Key <em>x<\/em>-value<\/strong><br>1. <strong>For <em>x<\/em> = \u22126:<\/strong><br><em>f<\/em>(\u22126) = (\u22126 + 6)(\u22126 + 5)(\u22126 \u2212 4)<br><em>f<\/em>(-6) = (0)(\u22121)(\u221210)<br><em>f<\/em>(-6) = 0 x -1 x -10<br><em>f<\/em>(-6) = 0<br>So, the value of <em>f<\/em>(-6) is equal to 0.<br><em>y<\/em> = <em>f<\/em>(\u22126) \u2212 3<br><em>y<\/em> = 0 \u2212 3<br><em>y<\/em> = \u22123<br><br>2. <strong>For <em>x<\/em> = \u22125:<\/strong><br><em>f<\/em>(\u22125) = (\u22125 + 6)(\u22125 + 5)(\u22125 \u2212 4)<br><em>f<\/em>(-5) = (1)(0)(\u22129)<br>f(-5) = 1 x 0 x -9<br><em>f<\/em>(-5) = 0<br>So, the value of <em>f<\/em>(-5) is equal to 0.<br><em>y<\/em> = <em>f<\/em>(\u22125) \u2212 3<br><em>y<\/em> = 0 \u2212 3<br><em>y<\/em> = \u22123<br><br>3. <strong>For <em>x<\/em> = 4:<\/strong><br><em>f<\/em>(4) = (4 + 6)(4 + 5)(4 \u2212 4)<br><em>f<\/em>(4) = (10)(9)(0)<br><em>f<\/em>(4) = 10 x 9 x 0<br><em>f<\/em>(4) = 0<br>So, <em>f<\/em>(4) is equal to 0.<br><em>y<\/em> = <em>f<\/em>(4) \u2212 3<br><em>y<\/em> = 0 \u2212 3<br><strong><em>y<\/em> = \u22123<\/strong><br><br>Hence, if:<br><em>x<\/em> = -6 then <em>y<\/em> is -3.<br><em>x<\/em> = -5 then <em>y<\/em> is -3.<br><em>x<\/em> = 4 then <em>y<\/em> is -3.<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">x<\/th><th class=\"has-text-align-center\" data-align=\"center\">y<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">-6<\/td><td class=\"has-text-align-center\" data-align=\"center\">-3<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">-5<\/td><td class=\"has-text-align-center\" data-align=\"center\">-3<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">4<\/td><td class=\"has-text-align-center\" data-align=\"center\">-3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Final Answer: Option B.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>17th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> The regular price of a shirt at a store is $11.70. The sale price of the shirt is 80% less than the regular price, and the sale price is 30% greater than the store\u2019s cost for the shirt. What was the store\u2019s cost, in dollars, for the shirt?<br>A) 1.7<br>B) 1.3<br>C) 1.8<br>D) 2.3<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice C<\/strong> is the correct option. The correct answer is 1.8. It\u2019s given that the regular price of a shirt at a store is $11.70, and the sale price of the shirt is 80% less than the regular price. It follows that the sale price of the shirt is $11.70(1 &#8211; 80\/100), or $11.70(1 &#8211; 0.8), which is equivalent to $2.34. It\u2019s also given that the sale price of the shirt is 30% greater than the store\u2019s cost for the shirt. Let <em>x<\/em> represent the store\u2019s cost for the shirt. It follows that 2.34 = (1 + 30\/100)<em>x<\/em>, or 2.34 = 1.3<em>x<\/em>. Dividing both sides of this equation by 1.3 yields <em>x<\/em> = 1.80. Therefore, the store\u2019s cost, in dollars, for the shirt is 1.80. Note that 1.8 and 9\/5 are examples of ways to enter a correct answer.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem Analysis:<\/strong><br>We are tasked with finding the store&#8217;s cost for a shirt when the following conditions are provided:<br>1. The regular price of the shirt is <strong>$11.70<\/strong>.<br>2. The <strong>sale price<\/strong> of the shirt is <strong>80% less than the regular price<\/strong>.<br>3. The <strong>sale price<\/strong> is <strong>30% greater than the store&#8217;s cost<\/strong> for the shirt.<br>We need to calculate the store&#8217;s cost.<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Calculate the Sale Price of the Shirt<\/strong><br>The sale price is <strong>80% less than the regular price<\/strong>. This means we subtract 80% of the regular price from the regular price:<br>Sale&nbsp;Price = Regular&nbsp;Price \u2212 0.8 <strong>\u22c5<\/strong> Regular&nbsp;Price<br>Substitute Regular&nbsp;Price = 11.70:<br>Sale&nbsp;Price = 11.70 \u2212 0.8 <strong>\u22c5<\/strong> 11.70<br>First, calculate 0.8 <strong>\u22c5<\/strong> 11.70<br>0.8 <strong>\u22c5<\/strong> 11.70 = 9.36<br>Now subtract:<br>Sale&nbsp;Price = 11.70 \u2212 9.36 = 2.34<br>The <strong>sale price is $2.34<\/strong>.<br><br><strong>Step 2: Relate the Sale Price to the Store&#8217;s Cost<\/strong><br>The sale price is <strong>30% greater than the store&#8217;s cost<\/strong>. Let the store&#8217;s cost be <em>C<\/em>. The sale price is given by:<br>Sale&nbsp;Price = <em>C<\/em> + 0.3 <strong>\u22c5<\/strong> C<br>Factor <em>C<\/em> from the right-hand side:<br>Sale&nbsp;Price = 1.3 <strong>\u22c5<\/strong> <em>C<\/em><br>Substitute Sale&nbsp;Price = 2.34<br>2.34 = 1.3 <strong>\u22c5<\/strong> <em>C<\/em><br><br><strong>Step 3: Solve for the Store&#8217;s Cost<\/strong><br>To find <em>C<\/em>, divide both sides of the equation by 1.3:<br><em>C<\/em> = 2.34\/1.3<br>Perform the division: C = 1.8<br>The <strong>store&#8217;s cost is $1.80<\/strong>.<br><br><strong>Verification:<\/strong><br>1. <strong>Check the Sale Price:<\/strong> If the store&#8217;s cost is $1.80, then the sale price (30% greater than the cost) is:<br>Sale&nbsp;Price = 1.3 <strong>\u22c5 <\/strong>1.80 = 2.34<br>This matches the given sale price, so the cost calculation is correct.<br>2. <strong>Check the Regular Price Reduction:<\/strong> The sale price of $2.34 is indeed 80% less than the regular price of $11.70:<br>80% of Regular Price = 0.8 <strong>\u22c5<\/strong> 11.70 = 9.36<br>Regular Price &#8211; 80% Reduction = 11.70 &#8211; 9.36 = 2.34<br>This confirms the correctness of the sale price.<br><br><strong>Final Answer:<\/strong> The store&#8217;s cost for the shirt is: <strong>1.80.\u200b<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>18th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> A certain town has an area of 4.36 square miles. What is the area, in <span style=\"text-decoration: underline\">square yards<\/span>, of this town? (1 mile = 1,760 yards)<br>A) 404<br>B) 7,674<br>C) 710,459<br>D) 13,505,536<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice D<\/strong> is correct. Since the number of yards in 1 mile is 1,760, the number of square yards in 1 square mile is (1,760)(1,760) = 3,097,600. Therefore, if the area of the town is 4.36 square miles, it is 4.36(3,097,600) = 13,505,536, in square yards.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from dividing the number of yards in a mile by the square mileage of the town.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from multiplying the number of yards in a mile by the square mileage of the town.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from dividing the number of square yards in a square mile by the square mileage of the town.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem Analysis:<\/strong><br>We need to convert the given area of the town from square miles to square yards. The given information is:<br>~ <strong>Area of the town:<\/strong> 4.36\u2009square&nbsp;miles<br>~ <strong>Conversion factor:<\/strong> 1\u2009mile = 1,760\u2009yards.<br>We know that to convert square miles to square yards, we need to <strong>square the conversion factor<\/strong> for miles to yards.<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Conversion Factor from Miles to Yards<\/strong><br>The given conversion is:<br>1\u2009mile = 1,760\u2009yards.<br>To convert <strong>square miles<\/strong> to <strong>square yards<\/strong>, square the conversion factor:(1\u2009mile)<sup>2<\/sup> = (1,760\u2009yards)<sup>2<\/sup>.<br>Perform the squaring:<br>1\u2009square&nbsp;mile = 1,760 \u00d7 1,760 = 3,097,600\u2009square&nbsp;yards.<br><br><strong>Step 2: Convert 4.36 Square Miles to Square Yards<\/strong><br>We are given the area of the town as 4.36\u2009square&nbsp;miles. Multiply this by the conversion factor:<br>Area&nbsp;in&nbsp;square&nbsp;yards = 4.36 \u00d7 3,097,600.<br>Perform the multiplication:<br>4.36 \u00d7 3,097,600 = 13,505,536\u2009square&nbsp;yards.<br><br><strong>Verification:<\/strong><br>Recompute the multiplication step for accuracy:<br>~ Multiply 4.36 by 3,097,600:<br>4.36 \u00d7 3,097,600 = 13,505,536.<br>The computation is correct.<br><br><strong>Final Answer:<\/strong> The area of the town is: <strong>13,505,536\u2009square&nbsp;yards.\u200b<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>19th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th class=\"has-text-align-center\" data-align=\"center\">x<\/th><th class=\"has-text-align-center\" data-align=\"center\">y<\/th><\/tr><\/thead><tbody><tr><td class=\"has-text-align-center\" data-align=\"center\">18<\/td><td class=\"has-text-align-center\" data-align=\"center\">130<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">23<\/td><td class=\"has-text-align-center\" data-align=\"center\">160<\/td><\/tr><tr><td class=\"has-text-align-center\" data-align=\"center\">26<\/td><td class=\"has-text-align-center\" data-align=\"center\">178<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> For line <em>h<\/em>, the table shows three values of <em>x<\/em> and their corresponding values of <em>y<\/em>. Line <em>k<\/em> is the result of translating line <em>h<\/em> down 5 units in the <em>xy<\/em>-plane. What is the <em>x<\/em>-intercept of line <em>k<\/em>?<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"97\" height=\"196\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-50.png\" alt=\"SAT Question and answer options and solutions\" class=\"wp-image-5072\"\/><\/figure>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice D<\/strong> is the correct answer.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"648\" height=\"307\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-46.png\" alt=\"Answers of the sat math problems and lessons\" class=\"wp-image-5052\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-46.png 648w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-46-300x142.png 300w\" sizes=\"auto, (max-width: 648px) 100vw, 648px\" \/><\/figure>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Find the equation of line <em>h<\/em><\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"604\" height=\"597\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-51.png\" alt=\"SAT Math Free Lessons and Learning\" class=\"wp-image-5076\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-51.png 604w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-51-300x297.png 300w\" sizes=\"auto, (max-width: 604px) 100vw, 604px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Step 2: Find the <em>y<\/em>-intercept (<em>b<\/em>) of line <em>h<\/em><\/strong><br>We substitute <em>m<\/em> = 6 and one point, such as (18, 130), into <em>y<\/em> = <em>mx<\/em> + <em>b<\/em>:<br>130 = 6(18) + <em>b<\/em>.<br>Simplify:<br>130 = 108 + <em>b<\/em>.<br>Solve for <em>b<\/em>:<br><em>b<\/em> = 130 \u2212 108<br><em>b<\/em> = 22.<br>The equation of line <em>h<\/em> is:<br><em>y<\/em> = 6<em>x<\/em> + 22.<br><br><strong>Step 3: Equation of line <em>k<\/em><\/strong><br>Line <em>k<\/em> is obtained by <strong>translating line <em>h<\/em> down 5 units<\/strong>. This means we subtract 5 from the <em>y<\/em>-values:<br><em>y<\/em> = 6<em>x<\/em> + 22 \u2212 5<br><em>y<\/em> = 6<em>x<\/em> + 17.<br>The equation of line <em>k<\/em> is:<br><em>y<\/em> = 6<em>x<\/em> + 17.<br><br><strong>Step 4: Find the x-intercept of line <em>k<\/em><\/strong><br>The <em>x<\/em>-intercept occurs when <em>y<\/em> = 0. Set <em>y<\/em> = 0 in the equation of line <em>k<\/em>:<br>0 = 6<em>x<\/em> + 17.<br>Solve for <em>x<\/em>:<br>6<em>x<\/em> = \u221217.<br><em>x<\/em> = \u221217\/6.<br>The <em>x<\/em>-intercept is: (\u221217\/6, 0).<br><br><strong>Verification:<\/strong><br>The slope <em>m<\/em> = 6 and <em>y<\/em>-intercept translation were computed correctly. Substituting values confirms that <em>x<\/em> = \u221217\/6 satisfies the equation.<br><br><strong>Final Answer: D)&nbsp;(\u221217\/6, 0).<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>20th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> In the <em>xy<\/em>-plane, the graph of the equation <em>y<\/em> = \u2212<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> \u2212 100 intersects the line <em>y<\/em> = <em>c<\/em> at exactly one point. What is the value of <em>c<\/em>?<br>A) \u2212 481\/4<br>B) \u2212100<br>C) \u2212 319\/4<br>D) \u2212 9\/2<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice C<\/strong> is correct. In the <em>xy<\/em>-plane, the graph of the line <em>y<\/em> = <em>c<\/em> is a horizontal line that crosses the <em>y<\/em>-axis at <em>y<\/em> = <em>c<\/em> and the graph of the quadratic equation <em>y<\/em> = &#8211;<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> &#8211; 100 is a parabola. A parabola can intersect a horizontal line at exactly one point only at its vertex. Therefore, the value of <em>c<\/em> should be equal to the <em>y<\/em>-coordinate of the vertex of the graph of the given equation. For a quadratic equation in vertex form, <em>y<\/em> = <em>a<\/em>(<em>x<\/em> &#8211; <em>h<\/em>)<sup>2<\/sup> + <em>k<\/em>, the vertex of its graph in the <em>xy<\/em>-plane is (<em>h<\/em>, <em>k<\/em>). The given quadratic equation, <em>y<\/em> = &#8211;<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> &#8211; 100, can be rewritten as<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"626\" height=\"83\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-47.png\" alt=\"Explanation of the SAT test (Math Solutions)\" class=\"wp-image-5054\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-47.png 626w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-47-300x40.png 300w\" sizes=\"auto, (max-width: 626px) 100vw, 626px\" \/><\/figure>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Problem Analysis:<\/strong><br>We are tasked with finding the value of <em>c<\/em> such that the graph of the quadratic equation <em>y<\/em> = \u2212<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> \u2212 100 intersects the horizontal line <em>y<\/em> = <em>c<\/em> <strong>at exactly one point<\/strong>. This condition implies that the quadratic equation has exactly <strong>one solution<\/strong>, meaning the discriminant (\u0394) of the quadratic equation must be <strong>0<\/strong>.<br><br><strong>Step-by-Step Solution:<\/strong><br><strong>Step 1: Set the two equations equal to each other<\/strong><br>The line <em>y<\/em> = <em>c<\/em> intersects the parabola <em>y<\/em> = \u2212<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> \u2212 100 when:<br>\u2212<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> \u2212 100 = <em>c<\/em>.<br>Rearrange the equation to form a standard quadratic equation:<br>\u2212<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> \u2212 100 \u2212 <em>c<\/em> = 0.<br>Simplify:<br>\u2212<em>x<\/em><sup>2<\/sup> + 9<em>x<\/em> + (\u2212100 \u2212 <em>c<\/em>) = 0.<br>This equation can also be written as:<br><em>x<\/em><sup>2<\/sup> \u2212 9<em>x<\/em> + (100 + <em>c<\/em>) = 0 (after&nbsp;multiplying&nbsp;through&nbsp;by&nbsp;-1).<br><br><strong>Step 2: Apply the discriminant condition<\/strong><br>For a quadratic equation of the form <em>ax<\/em><sup>2<\/sup> + <em>bx<\/em> + <em>c<\/em> = 0, the discriminant is given by:<br>\u0394 = <em>b<\/em><sup>2<\/sup> \u2212 4<em>ac<\/em>.<br>Here, the coefficients are:<br><em>a<\/em> = 1,<br><em>b<\/em> = \u22129, and<br><em>c<\/em> = 100 + <em>c<\/em>.<br>The quadratic equation has <strong>exactly one solution<\/strong> when <strong>\u0394<\/strong> = 0. Set the discriminant equal to 0:<br>\u0394 = (\u22129)<sup>2<\/sup> \u2212 4(1)(100 + <em>c<\/em>) = 0.<br>Simplify: 81 \u2212 4(100 + <em>c<\/em>) = 0.<br><br><strong>Step 3: Solve for <em>c<\/em><\/strong><br>Distribute: 81 \u2212 400 \u2212 4<em>c<\/em> = 0.<br>Simplify: \u2212319 \u2212 4<em>c<\/em> = 0.<br>Solve for <em>c<\/em>:<br>4<em>c<\/em> = \u2212319.<br><em>c<\/em> = \u2212319\/4.<br><br><strong>Final Answer: C)&nbsp;\u2212319\/4.\u200b\u200b<\/strong><br><br><strong>Verification:<\/strong><br>Substitute <em>c<\/em> = \u2212319\/4\u200b into the quadratic equation <em>x<\/em><sup>2<\/sup> \u2212 9<em>x<\/em> + (100 + <em>c<\/em>) = 0. The discriminant calculation will confirm that <strong>\u0394<\/strong> = 0, verifying that the parabola intersects the line at exactly one point.<br><br><strong>Final Answer: C)&nbsp;\u2212319\/4.\u200b\u200b<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>21th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>2<em>x<\/em> + 3<em>y<\/em> = 7<br>10<em>x<\/em> + 15<em>y<\/em> = 35<\/strong><br><strong>Question:<\/strong> For each real number <em>r<\/em>, which of the following points lies on the graph of each equation in the <em>xy<\/em>-plane for the given system?<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"138\" height=\"200\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-52.png\" alt=\"Math question and answer solutions and explanations\" class=\"wp-image-5085\"\/><\/figure>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice B<\/strong> is correct.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"644\" height=\"310\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-48.png\" alt=\"SAT Math Solutions and Lessons for free\" class=\"wp-image-5056\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-48.png 644w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-48-300x144.png 300w\" sizes=\"auto, (max-width: 644px) 100vw, 644px\" \/><\/figure>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice A is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">It is an option-based question, which means without implying, the given equation of the question, into the options, you cannot find the correct one.<br><strong>Step 1: The System of Equations<\/strong><br>The system of equations is:<br>2<em>x<\/em> + 3<em>y<\/em> = 7<br>10<em>x<\/em> + 15<em>y<\/em> = 35<br>The second equation is a multiple of the first [10<em>x<\/em> + 15<em>y<\/em> = 5(2<em>x<\/em> + 3<em>y<\/em>)], so we only need to check which option satisfies 2<em>x<\/em> + 3<em>y<\/em> = 7. You can choose the first or second equation from above, it doesn&#8217;t matter but we are against the time so let&#8217;s pick that saves time, equation first.<br><br><strong>Step 2: Substituting Each Option into 2<em>x<\/em> + 3<em>y<\/em> = 7<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1175\" height=\"643\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-53.png\" alt=\"SAT Free Preparation (Question and answer solutions)\" class=\"wp-image-5089\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-53.png 1175w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-53-300x164.png 300w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-53-1024x560.png 1024w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-53-768x420.png 768w\" sizes=\"auto, (max-width: 1175px) 100vw, 1175px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">Let&#8217;s use the same substitute for Options C and D, even though we have our correct option, we can still solve the complete question for better preparation.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1290\" height=\"594\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-54.png\" alt=\"SAT Free Preparation (Question and answer Explanations)\" class=\"wp-image-5091\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-54.png 1290w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-54-300x138.png 300w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-54-1024x472.png 1024w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-54-768x354.png 768w\" sizes=\"auto, (max-width: 1290px) 100vw, 1290px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">6. we have got -5<em>r<\/em> + 21 = 14 of Option D: Solve for <em>r<\/em><br>-5<em>r<\/em> = 14 &#8211; 21<br>-5<em>r<\/em> = -7 (we can cut the minus sign from both sides.)<br><strong><em>r<\/em> = 7\/5.<\/strong><br><br>Now, we have the <em>r<\/em> value from all 4 Options. Let&#8217;s put them in the given equation: 2<em>x<\/em> + 3<em>y <\/em>= 7 (If you use this equation to solve the question then <em>r<\/em> = 7. Whichever option gives you <em>r<\/em> = 7 is the correct option. But if you use the second equation 10<em>x<\/em> + 15<em>y<\/em> = 35, then <em>r<\/em> = 35.)<br>The correct option is <strong>Option B<\/strong>.<br>-3<em>r<\/em> + 7 + 3<em>r<\/em><br>-3<em>r<\/em> + 3<em>r<\/em> + 7 (Both 3<em>r<\/em> cut one another)<br><strong>7.<\/strong><br><br><strong>Final Answer: Option B.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n\n\n<div class=\"wp-block-coblocks-accordion-item\"><details><summary class=\"wp-block-coblocks-accordion-item__title\"><strong>22th Question<\/strong><\/summary><div class=\"wp-block-coblocks-accordion-item__content\">\n<p class=\"is-style-warning\" style=\"font-size:0.9em\"><strong>Question:<\/strong> <\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"308\" height=\"96\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-55.png\" alt=\"SAT Math question and answer preparation and solutions\" class=\"wp-image-5100\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-55.png 308w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-55-300x94.png 300w\" sizes=\"auto, (max-width: 308px) 100vw, 308px\" \/><\/figure>\n\n\n\n<p class=\"is-style-warning\" style=\"font-size:0.9em\">A) 104<br>B) 624<br>C) 180<br>D) 60<br>[Type-Based Answer: In the final exam, you will type the answer rather than choose from options.]<\/p>\n\n\n\n<p class=\"is-style-success\" style=\"font-size:0.9em\"><strong>Choice A<\/strong> is correct. The correct answer is 104. An equilateral triangle is a triangle in which all three sides have the same length and all three angles have a measure of 60<sup>o<\/sup>. The height of the triangle, k root of 3, is the length of the altitude from one vertex. The altitude divides the equilateral triangle into two congruent 30-60-90 right triangles, where the altitude is the side across from the 60<sup>o<\/sup> angle in each 30-60-90 right triangle. Since the altitude has a length of k root of 3, it follows from the properties of 30-60-90 right triangles that the side across from each 30<sup>o<\/sup> angle has a length of <em>k<\/em> and each hypotenuse has a length of 2<em>k<\/em>. In this case, the hypotenuse of each 30-60-90 right triangle is a side of the equilateral triangle; therefore, each side length of the equilateral triangle is 2<em>k<\/em>. The perimeter of a triangle is the sum of the lengths of each side. It\u2019s given that the perimeter of the<br>equilateral triangle is 624; therefore, 2<em>k<\/em> + 2<em>k<\/em> + 2<em>k<\/em> = 624, or 6<em>k<\/em> = 624. Dividing both sides of this equation by 6 yields <em>k<\/em> =104.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice B is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice C is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-error\" style=\"font-size:0.9em\">Choice D is incorrect and may result from conceptual or calculation errors.<\/p>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\">Learn to use the Desmos calculator, so this question will become easy for you. Let&#8217;s start!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1288\" height=\"604\" src=\"https:\/\/us.mrenglishkj.com\/sat\/sat\/wp-content\/uploads\/2024\/12\/image-56.png\" alt=\"Math Problems solutions and explanations (SAT Free Preparation)\" class=\"wp-image-5102\" srcset=\"https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-56.png 1288w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-56-300x141.png 300w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-56-1024x480.png 1024w, https:\/\/us.mrenglishkj.com\/sat\/wp-content\/uploads\/sites\/2\/2024\/12\/image-56-768x360.png 768w\" sizes=\"auto, (max-width: 1288px) 100vw, 1288px\" \/><\/figure>\n\n\n\n<p class=\"is-style-info\" style=\"font-size:0.9em\"><strong>Final Answer: The value of <em>k<\/em> is: 4.<\/strong><\/p>\n<\/div><\/details><\/div>\n<\/div>\n\n\n\n<div style=\"height:70px\" aria-hidden=\"true\" class=\"wp-block-spacer\"><\/div>\n\n\n\n<p>Did you try all the features and get comfortable using them? You should work on using the calculator and seeing references and directions. So be prepared for everything before taking the final SAT exam. The explanation of answers makes it easy to learn and progress. You must try to work on your speed and spend less time on the beginning and more on the later questions. This is the 6th Practice Test of SAT Math Module 1st.<\/p>\n\n\n\n<p>Either you can take the 7th Practice Test of SAT Math or the 6th Practice Test of SAT Math Module 2nd.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><a href=\"https:\/\/us.mrenglishkj.com\/sat\/sat-math-test-6-module-2nd-lessons\/\" target=\"_blank\" rel=\"noopener\" title=\"\">SAT Test 6th (Math Module 2nd)<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/us.mrenglishkj.com\/sat\/sat-math-test-7-module-1st-prepare-for-sat\/\" target=\"_blank\" rel=\"noopener\" title=\"\">SAT Test 7th (Math Module 1st)<\/a><\/li>\n\n\n\n<li><a href=\"https:\/\/us.mrenglishkj.com\/sat\/sat-test-6-reading-and-writing-1st-module\/\" target=\"_blank\" rel=\"noopener\" title=\"\">SAT Test 6th (Reading and Writing Module 1st)<\/a><\/li>\n<\/ul>\n\n\n\n<p>The best way to become a master in Math is to find the correct answer and understand why other options are incorrect. I wish you luck in your bright career.<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>SAT Math Test 6 Module 1st Real Exam Question Tips &#038; Tricks: SAT Math Preparation for free, learn SATs by taking the test and studying the answer solutions. Solve SAT questions and answers with real-time experience<\/p>\n","protected":false},"author":1,"featured_media":4909,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"googlesitekit_rrm_CAowmvTFDA:productID":"","_coblocks_attr":"","_coblocks_dimensions":"","_coblocks_responsive_height":"","_coblocks_accordion_ie_support":"","footnotes":""},"categories":[12],"tags":[27,28],"class_list":["post-4907","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-1st-module","tag-sat-math","tag-sat-module-1st"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/posts\/4907","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/comments?post=4907"}],"version-history":[{"count":1,"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/posts\/4907\/revisions"}],"predecessor-version":[{"id":8826,"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/posts\/4907\/revisions\/8826"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/media\/4909"}],"wp:attachment":[{"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/media?parent=4907"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/categories?post=4907"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/us.mrenglishkj.com\/sat\/wp-json\/wp\/v2\/tags?post=4907"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}