Answer:
x=20
Step-by-step explanation:
The Pythagorean theorem is
a^2 +b^2 = c^2
where a and b are the legs and c is the hypotenuse.
The legs are x and 15 and the hypotenuse is 25
x^2 + 15^2 = 25^2
x^2 +225 = 625
Subtract 225 from each side
x^2 +225-225 = 625-225
x^2 = 400
Take the square root of each side
sqrt(x^2) = sqrt(400)
x = 20
The two labeled angles are alternate interior angles, and as such, they are the same.
From this result you can build the equation

and solve it for x: subtract 13x from both sides to get

and add 2 to both sides to get

Check: if we plug the value we found we have

So the angles are actually the same, as requested.
Answer:
A number and its additive inverse are always the same distance from zero, and so they have the same absolute value.
Step-by-step explanation:
Hope this helps!!
Answer:
I think it would be C.
Step-by-step explanation:
One is positive Four is negative so is 1+(-4)
Answer:

Step-by-step explanation:
