Question

Y is partly constant and partly varies as the square of x when y=16,x=2 and when y=24 x=3
find the relation between x and y
find y when x=4

Answer 1

Given that y is partly constant and partly varies as the square of x, we can write the relation as:

y = kx^2 + c

where k is the factor by which y varies as the square of x and c is the constant part of y.

Using the given values, when y = 16 and x = 2:
16 = k * 2^2 + c
16 = 4k + c

And when y = 24 and x = 3:
24 = k * 3^2 + c
24 = 9k + c

Subtracting the first equation from the second gives:
24 - 16 = 9k + c - (4k + c)
8 = 5k
k = 8/5

Substitute back into the first equation to find c:
16 = 4 * (8/5) + c
16 = 32/5 + c
16 - 32/5 = c
80/5 - 32/5 = c
48/5 = c

Therefore, the relation between x and y is:
y = (8/5)x^2 + 48/5

To find y when x = 4:
y = (8/5) * 4^2 + 48/5
y = (8/5) * 16 + 48/5
y = 128/5 + 48/5
y = 176/5
y = 35.2

Therefore, when x = 4, y = 35.2.