A triangle has three sides of the following side lengths: 7, 10, and x^2. What are all of the positive integer values of x such that the triangle exists? Separate your answers using commas and express them in increasing order.

x = {2,3,4}    (if x can only be positive whole numbers)

Step-by-step explanation:

For a triangle exists, the side lengths of the triangle must be such that the sum of the two shorter sides must be greater than the third side.

This also is equivalent to any two sides must have a sum greater than the third side.

So

7+10 > x^2, => x^2 < 17 => x < sqrt(17)     (maximum)

7+x^2 > 10, => x^2 >3 => x > sqrt(3)

Therefore

sqrt(3) < x < sqrt(17)

If x must be an integer,

2< x < 4, or x = {2,3,4}

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.