URGENT! Please help me solve this question with full solutions

C) 2/3

Explanation:

Let's cross multiply and then get everything to one side

x/(2x-1) = (2x+1)/(x+2)

x(x+2) = (2x-1)(2x+1)

x^2+2x = 4x^2-1

0 = 4x^2-1-x^2-2x

0 = 3x^2-2x-1

3x^2-2x-1 = 0 see note below

x^2-(2/3)x-1/3 = 0

In that last step, I divided everything by the leading coefficient 3. That way, the leading coefficient becomes 1.

It turns out that anything in the form x^2-px+q has roots m and n such that

p = m+n

q = m*n

In short, the roots add to the negative of the middle coefficient, and they also multiply to the constant term. Again this only applies to quadratics where the leading coefficient is 1.

Comparing x^2-(2/3)x-1/3 with the form x^2-px+q, we see that

p = 2/3

q = -1/3

Therefore, p = m+n = 2/3

If you're curious why this works, it's because (x+a)(x+b) expands out to x^2+(a+b)x+ab, and note how the roots here are x = -a and x = -b.

Note: Another route would be to solve 3x^2-2x-1 = 0 for x using any method you prefer (I recommend the quadratic formula). You would get the two solutions x = -1/3 and x = 1. They add to (-1/3)+1 = (-1/3) + (3/3) = (-1+3)/3 = 2/3.

goog may be able to you find that answer

answer: 75°

step-by-step explanation:

∠esk is the supplement of ∠edk, so is 100°. ∠esd = 1/2·arc ed, so is 25°.

since ∠esk = ∠esd + ∠dsk

you have 100° = 25° + ∠dsk . .

fill in known values 75° = ∠dsk . . . . .

subtract 25°

step-by-step explanation: