Can someone me and don’t send random answers because is due tomorrow you

First find the hypotenuse of the triangle with side lengths sqrt 2 and sqrt 2:

(sqrt2)^2 + (sqrt2)^2 = x^2

x^2 = 4

x = 2

Since x is one of the legs of the triangle with side lengths y and 1, we can use the Pythagorean Theorem again:

1^2 + 2^2 = y^2

y^2 = 5

y = sqrt5, or choice (B).

The right triangle with two sides of length sqrt(2) is a 45-45-90 triangle.

The hypotenuse is sqrt(2) times the length of the leg, so the hypotenuse is 2.

The hypotenuse of that triangle is a leg of the other right triangle. The second right triangle has a leg of length 2, and leg of length 1. We can use the Pythagorean theorem to find y.

y^2 = a^2 + b^2

y^2 = 1^2 + 2^2

y^2 = 1 + 4

y^2 = 5

y = sqrt(5)

y = sqrt(5)

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

answer: download photomath it's legit.

step-by-step explanation: