Question

Find the constant of variation for the quadratic variation.
x:2,3,4,5,6
y:32,72,128,200,288
8
16
24
40

Answer 1

\(\bf \begin{array}{cccllll}
\textit{something }&\textit{varies directly to }&\textit{something else}\\ \quad \\
\textit{something }&={\color{purple}{ \textit{some value }}}&\textit{something else}\\ \quad \\
y&={\color{purple}{ n}}&x&\implies y={\color{purple}{ n}}x^2
\end{array}\\ \quad \\
\begin{array}{ccccccc}
x&&2&3&{\color{brown}{ 4}}&5&6\\
y&&32&72&{\color{blue}{ 128}}&200&288
\end{array}\qquad {\color{blue}{ y}}={\color{purple}{ n}}{\color{brown}{ x}}^2
\\ \quad \\
{\color{purple}{ n}}=\textit{constant of variation}\)

Answer 2

tis \(\large x^2\) since is a quadratic, thus

Answer 3

so, just solve for "n"

Answer 4

ok is the answer d?

Answer 5

dunno, what did you get for "n"?

Answer 6

Well I did 128n\[=n4^{2}\]

Answer 7

then 128=n16

Answer 8

128n?

Answer 9

i got 112?

Answer 10

\(\bf \begin{array}{ccccccc}
x&&2&3&{\color{brown}{ 4}}&5&6\\
y&&32&72&{\color{blue}{ 128}}&200&288
\end{array}\qquad {\color{blue}{ y}}={\color{purple}{ n}}{\color{brown}{ x}}^2
\\ \quad \\
{\color{purple}{ n}}=\textit{constant of variation}\implies {\color{blue}{ 128}}={\color{purple}{ n}}{\color{brown}{ 4}}^2
\\ \quad \\
\implies 128=16n\implies \cfrac{\cancel{128}}{\cancel{16}}=n\)

Answer 11

8?

Answer 12

yeap

Answer 13

THANKS

Answer 14

yw