Solve for x in the right triangle. i'm stuck on the early units in my final it would mean the world to me if someone could explain this. in advance.

Wel have to use the Pythagorean theorem (a^2) + (b^2) = (c^2)

look at the top triangle inside the big triangle. we know it's 2 sides but not the hypotenuse. one side would be 25 - 9 = 16 and other side is x. So so from theorem we know
16^2 + x^2 = h^2

with bottom triangle we can see the same thing but with 9 and x as the sides so:
9^2 + x ^2 = H^2 (capital H to distinguish them)

we also know that in the main triangle:
h^2 + H^2 = 25^2

now all we have to do is find a way to use these 3 equations to get the answer

we know what h^2 and H^2 both equal and can plug them into equation in orsee to get rid of one and solve for the other. So we use:
9^2 + x ^2 = H^2

and plug it into
h^2 + H^2 = 25^2

and get
h^2 + 9^2 + x ^2 = 25^2

solve for h^2
25^2 - x^2 - 9^2 = h^2

now plug what we got for h^2 into:
16^2 + x^2 = h^2

and get:
16^2 + x^2 = 25^2 - x^2 - 9^2

2x^2 = 25^2 - 16^2 - 9^2
2x^2 = 288
x^2 = 144
x = 12

there is your answer

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.