The image presents three math problems. (i) Find the value of $k$ given $y = kx^2$ and $y = 4$ when $x = 1$. (ii) Find the value of $y$ when $x = 5$, using the $k$ found in problem (i). Also, find the value of $x$ when $y = 36$, using the $k$ found in problem (i). (iii) Find the value of $k$ given that $y$ varies directly as $x$ and inversely as $z$, $y=4$ when $x=8$ and $z=2$. The relationship can be represented as $y = k\frac{x}{z}$.

1. Problem Description

The image presents three math problems.

(i) Find the value of $k$ given $y = kx^2$ and $y = 4$ when $x = 1$.

(ii) Find the value of $y$ when $x = 5$, using the $k$ found in problem (i). Also, find the value of $x$ when $y = 36$, using the $k$ found in problem (i).

(iii) Find the value of $k$ given that $y$ varies directly as $x$ and inversely as $z$, $y=4$ when $x=8$ and $z=2$. The relationship can be represented as $y = k\frac{x}{z}$.

2. Solution Steps

(i) Finding the value of $k$:

Given $y = kx^2$ and $y = 4$ when $x = 1$.

Substituting the values of $x$ and $y$ into the equation:

$4 = k(1)^2$

$4 = k(1)$

$4 = k$

$k = 4$

(ii) Finding $y$ when $x = 5$:

Using the equation $y = kx^2$ and the value $k = 4$.

Substituting $x = 5$ and $k = 4$ into the equation:

$y = 4(5)^2$

$y = 4(25)$

$y = 100$

Finding $x$ when $y = 36$:

Using the equation $y = kx^2$ and the value $k = 4$.

Substituting $y = 36$ and $k = 4$ into the equation:

$36 = 4x^2$

Divide both sides by 4:

$9 = x^2$

Taking the square root of both sides:

$x = \pm 3$

Since the image indicates $x=3$, we'll take the positive root.

$x = 3$

(iii) Finding the value of $k$:

Given $y$ varies directly as $x$ and inversely as $z$, so $y = k\frac{x}{z}$.

We are given $y = 4$ when $x = 8$ and $z = 2$.

Substituting these values into the equation:

$4 = k\frac{8}{2}$

$4 = k(4)$

Dividing both sides by 4:

$1 = k$

$k = 1$

3. Final Answer

(i) $k = 4$

(ii) $y = 100$ when $x = 5$, and $x = 3$ when $y = 36$

(iii) $k = 1$