# Maths

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#1
y is directly proportional to x^2.
When x = 3, then y = 36.
a) Express y in terms of x.

z is inversely proportional to x.
When x = 4, z = 2.
b) Show that z = c y^n , where c and n are numbers and c > 0.

You must find the values of c and n.
0
4 years ago
#2
(Original post by choiheesung97)
y is directly proportional to x^2.
When x = 3, then y = 36.
a) Express y in terms of x.

z is inversely proportional to x.
When x = 4, z = 2.
b) Show that z = c y^n , where c and n are numbers and c > 0.

You must find the values of c and n.
a) y = kx^2
36 = (3^2)k
Now you just need to find k and substitute it back into the equation in the first line of my working out.
Spoiler:
Show

k= 36/9 =4
y = 4x^2

b) z = k/x
2 = k/4
Now find k
Spoiler:
Show

k= 2 x 4 = 8

Rearrange and substitute the equations into each other until they are in the form z = cy^n
z= 8/x
x = 8/z
Spoiler:
Show

x^2 = 64/z^2
y= 4(64/z^2)
y= 256/z^2
z^2 = 256/y
z = square root (256/y)
z = 16/ (y^0.5)
z = 16 y^(-0.5)

c = 16
y = -0.5

I have probably make a silly mistake somewhere but I enjoyed this question 0
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