Question

Given that y varies directly to the square root of (x + 1), and that y = 1 when x = 8,

(a)
express y in terms of .x,
(b)
find the value of y when x = 3,
(c)
find the value of x when y = 5.​

Answer 1

a) y=1/3 sqrt(x+1)

b) 2/3

c) 224

Explanation:

a) y varies directly to sqrt(x+1)

means there is some constant k such that y=ksqrt(x+1)

y=1 when x=8 (plug this in and find k) 1=ksqrt(8+1)

1=ksqrt(9)

1=k(3)

1/3=k

So the equation is y=1/3 * sqrt(x+1)

a) y=1/3 *sqrt(x+1)

b) what is y when x=3 if you have y=1/3 * sqrt(x+1)

Plug in 3 for x

y=1/3 *sqrt(3+1)

y=1/3 *sqrt(4)

y=1/3 *2

y=2/3

b) 2/3

c) what is x when y=5

Plug in 5 for y

5=1/3 *sqrt(x+1)

multiply both sides by 3

15=sqrt(x+1)

square both sides

225=x+1

subtract 1 on both sides

224=x

c) 224