I need help with these equations before midnight! Must show work, thank you!

Question

What is the solution of (x+4)^2/3 = 16

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The radius r of a circle with area A is given by r=(A/p)^1/2

What is the radius of a circular pool with an area of 100p ft^2 ?

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Radical symbol --> x+5 = 3/2x-3 must solve equation by graphing

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Please help!

Answer 1

1) (x+4)^2/3 = 16

x+4 = 16^3/2 = 64

x = 60

2) r= (A/p)^1/2

r = (100p/p)^1/2 = √100 = 10

3) if you graph √(x+5) - 3/(2x-3), the x-intercept is at 2.064

Answer 2

(x+4)^2 = 16^3 = 4096

x+4 = sqr4096 = - or + 64

x =-68 + 60

(60+4)^2/3 = 64^2/3 = square root of 64^3 = sqr4096 = - or + 64

A=100pi sq. ft.

r = sqr(100pi/pi) = sqr100 = 10 feet radius

x+5 = (3/2)x -3

x/2 = 8

x = 16

x+5 is a straight line with slope = 1 and y intercept = 5

(3/2)x -3 is a straight line with slightly steeper slope = 3/2 and y intercept -3

they intersect at (16,21)