Question

II. Solve the problems involving variations. Show your complete solution. (3 points each) 1. If y varies directly as the square of x, and y = 32 when x = 2, find y when x = 5. 2. The force of attraction (F) between two opposite electrical charges varies inversely as the square of the distance (d) between them. If F = 18 when d = 10, find F when d = 15. 3. If y varies jointly as x and z, find y if x = 3, k = 6 and z = 9​

Answer 1

Answer:

see explanation

Step-by-step explanation:

(1)

y varies directly as x² then the equation relating them is

y = kx² ← k is the constant of variation

To find k use the condition y = 32 when x = 2

32 = k × 2² = 4k ( divide both sides by 4 )

8 = k

y = 8x² ← equation of variation

When x = 5 , then

y = 8 × 5² = 8 × 25 = 200

(2)

Given F varies inversely as d² then the equation relating them is

F = $\frac{k}{d^2}$ ← k is the constant of variation

To find k use the condition F = 18 when d = 10

18 = $\frac{k}{10^2}$ = $\frac{k}{100}$ ( multiply both sides by 100 )

1800 = k

F = $\frac{1800}{d^2}$ ← equation of variation

When d = 15 , then

F = $\frac{1800}{15^2}$ = $\frac{1800}{225}$ = 8

(3)

y varies jointly as x and z then the equation relating them is

y = kxz ← k is the constant of variation

when x = 3, y = 6, z = 9 ,then

y = 6 × 3 × 9 = 162