The shorter sides of an acute triangle are x cm and 2x cm. the longest side of the trinalge is 15 cm. what is the smallest possible whole-number value of x ?

The smallest possible whole-number value of x is 7

Explanation:
Assume sides of the triangle are:
a = x
b = 2x
c = 15
We are given that c is the longest side
For the triangle to be acute:
c^2 < a^2 + b^2

Substitute with the values of a, b and c and solve for x as follows:
c^2 < a^2 + b^2
(15)^2 < (x)^2 + (2x)^2
225 < x^2 + 4x^2
225 < 5x^2
45 < x^2

For we will get the zeros, this means that we will solve for x^2 = 45:
x^2 = 45
x = + or - √45
This means that:
either x = 6.708
oR x = -6.708

We want the x^2 to be greater than 45.
This means that we want the x to be greater than 6.708
Therefore, the smallest possible whole number to satisfy this condition is 7.

Hope this helps :)

A. p = 2(l + w)p/2 - l = ww = p/2 - l b. while isolating the variable w in the equation, i first divided both sides by 2 and then i subtracted l from both sides. i hope this .