In a positive integers there are twenty whole numbers I hope this help
To get rid of

, you have to take the third root of both sides:
![\sqrt[3]{x^{3}} = \sqrt[3]{1}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B1%7D%20)
But that won't help you with understanding the problem. It is better to write

as a product of 2 polynomials:

From this we know, that

is the solution. Another solutions (complex roots) are the roots of quadratic equation.
Answer: y = 4
explanation:
y = 1/3(-3) + 5
y = (-3)/3 + 5
y = -1 + 5
y = 4
Complementary angles = 90
x+y =90
y=x+24
substitute this in to the equation
x+x+24=90
2x+24=90
subtract 24 from each side
2x = 66
divide by 2
x=33
y = 33+24
y=57
Answer: 33, 57
Try 1(18) / 2
Just multiply 1 by 18 and divide 18 by 2.
Then we have 9!