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1
The table below shows some corresponding values of
x and
(y-3)^(2) where
y is positive.

x
2
8
32

(y-3)^(2)
4
16
64

x
(i) Explain with evidence whether
x is directly proportional to
(y-3)^(2).
(ii) Express
x in terms of
y.
(iii) Find the value of
x when
y=1.2.
(iv) Find the value of
y when
x=72.

$1$ $\newline$ The table below shows some corresponding values of $x$ and $(y-3)^{2}$ where $y$ is positive. $\newline$ \begin{tabular}{|c|c|c|c|} $\newline$ \hline $x$ & $2$ & $8$ & $32$ \\ $\newline$ \hline $(y-3)^{2}$ & $4$ & $16$ & $64$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ $x$ $\newline$ (i) Explain with evidence whether $x$ is directly proportional to $(y-3)^{2}$ . $\newline$ (ii) Express $x$ in terms of $y$ . $\newline$ (iii) Find the value of $x$ when $(y-3)^{2}$ $1$ . $\newline$ (iv) Find the value of $y$ when $(y-3)^{2}$ $3$ .

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Write and solve direct variation equations

Full solution

Q.

1

\newline

The table below shows some corresponding values of

x

and

(y-3)^{2}

where

y

is positive.

\newline

\begin{tabular}{|c|c|c|c|}

\newline

\hline

x

&

2

&

8

&

32

\\

\newline

\hline

(y-3)^{2}

&

4

&

16

&

64

\\

\newline

\hline

\newline

\end{tabular}

\newline

x

\newline

(i) Explain with evidence whether

x

is directly proportional to

(y-3)^{2}

.

\newline

(ii) Express

x

in terms of

y

.

\newline

(iii) Find the value of

x

when

(y-3)^{2}

1

.

\newline

(iv) Find the value of

y

when

(y-3)^{2}

3

.

Check Ratios: (i) To check if $x$ is directly proportional to $(y-3)^{2}$ , look at the ratios of $x$ to $(y-3)^{2}$ . For $x=2$ , $(y-3)^{2}=4$ , ratio is $\frac{2}{4}=0.5$ . For $x=8$ , $(y-3)^{2}=16$ , ratio is $\frac{8}{16}=0.5$ . For $(y-3)^{2}$ $0$ , $(y-3)^{2}$ $1$ , ratio is $(y-3)^{2}$ $2$ . Since the ratios are constant, $x$ is directly proportional to $(y-3)^{2}$ .
Find Constant: (ii) Since $x$ is directly proportional to $(y-3)^{2}$ , we can write $x$ as a multiple of $(y-3)^{2}$ . Let's use the first set of values to find the constant of proportionality, $k$ . $x=k*(y-3)^{2}$ $2=k*4$ $k=\frac{2}{4}$ $k=0.5$ So, $x=0.5*(y-3)^{2}$ .
Substitute and Solve: (iii) To find $x$ when $y=1.2$ , substitute $y=1.2$ into $x=0.5\cdot(y-3)^{2}$ .
$x=0.5\cdot(1.2-3)^{2}$
$x=0.5\cdot(-1.8)^{2}$
$x=0.5\cdot3.24$
$x=1.62$

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