{"@context":"https://schema.org/","@type":"Quiz","name":"

Let the function $p$ be defined as $p(x)=\\frac{(x-c)^{2}+160}{2c}$ , where $c$ is a constant. If $p(c)=10$ , what is the value of $p(12)$ ?

","hasPart":{"@type":"Question","typicalAgeRange":"12-22","educationalAlignment":[{"@type":"AlignmentObject","alignmentType":"educationalSubject","targetName":"SAT"},{"@type":"AlignmentObject","alignmentType":"educationalLevel","targetName":"SAT","educationalFramework":"Common Core","targetUrl":"https://classnotes.xyz/mcqs/sat-math-advanced-math-multiple-choice-question-213"}],"about":{"@type":"Thing","name":"Math, Advanced Math, Duplicate"},"educationalLevel":"intermediate","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","assesses":["Math","Advanced Math","Duplicate"],"name":"

Let the function $p$ be defined as $p(x)=\\frac{(x-c)^{2}+160}{2c}$ , where $c$ is a constant. If $p(c)=10$ , what is the value of $p(12)$ ?

","comment":{"@type":"Comment","text":"

Choice D is correct. The value of $p(12)$ depends on the value of the constant $c$, so the value of $c$ must first bedetermined. It is given that $p(c) = 10$. Based on the definition of $p$, it follows that:

$p(c)=\\frac{(c-c)^{2}+160}{2c}=10$
$\\frac{160}{2c}=10$
$2c=16$
$c=8$

This means that $p(x)=\\frac{(x-8)^{2}+160}{16}$ for all values of $x$. Therefore:

$p(12)=\\frac{(12-8)^{2}+160}{16}$
$=\\frac{16+160}{16}$
$=11$

Choice A is incorrect. It is the value of $p(8)$, not $p(12)$. Choices B and C are incorrect. If one of these values were correct, then $x = 12$, and the selected value of $p(12)$ could be substituted into the equation to solve for $c$. However, the values of $c$ that result from choices B and C each result in $p(c) < 10$.

"},"encodingFormat":"text/markdown","text":"

Let the function $p$ be defined as $p(x)=\\frac{(x-c)^{2}+160}{2c}$ , where $c$ is a constant. If $p(c)=10$ , what is the value of $p(12)$ ?

","suggestedAnswer":[{"@type":"Answer","position":0,"encodingFormat":"text/html","text":"

10.00

"},{"@type":"Answer","position":1,"encodingFormat":"text/html","text":"

10.25

"},{"@type":"Answer","position":2,"encodingFormat":"text/html","text":"

10.75

"}],"acceptedAnswer":{"@type":"Answer","position":3,"encodingFormat":"text/html","text":"

11.00

","comment":{"@type":"Comment","text":"

$p(c)=\\frac{(c-c)^{2}+160}{2c}=10$
$\\frac{160}{2c}=10$
$2c=16$
$c=8$

This means that $p(x)=\\frac{(x-8)^{2}+160}{16}$ for all values of $x$. Therefore:

"},"answerExplanation":{"@type":"Comment","text":"

$p(c)=\\frac{(c-c)^{2}+160}{2c}=10$
$\\frac{160}{2c}=10$
$2c=16$
$c=8$

This means that $p(x)=\\frac{(x-8)^{2}+160}{16}$ for all values of $x$. Therefore:

"}}},"url":"https://classnotes.xyz/mcqs/sat-math-advanced-math-multiple-choice-question-213"}

Let the function $p$ be defined as $p(x)=\frac{(x-c)^{2}+160}{2c}$ , where $c$ is a constant. If $p(c)=10$ , what is the value of $p(12)$ ?