HELP PLEASE! Given that sin A=3/7, cos B=-2/5, and both AA and B are in quadrant II, find cos (A-B). Simplify to a single value and leave it in the form of a rational number.

First, recall that

cos(A - B) = cos(A) cos(B) + sin(A) sin(B)

so you just need to find cos(A) and sin(B).

Since both A and B end in the second quadrant, you know that

• cos(A) and cos(B) are both negative

• sin(A) and sin(B) are both positive

Then from the Pythagorean identity, you get

cos²(A) + sin²(A) = 1   ==>   cos(A) = -√(1 - sin²(A)) = -2√10/7

cos²(B) + sin²(B) = 1   ==>   sin(B) = +√(1 - cos²(B)) = √21/5

You'll end up with

cos(A - B) = (-2√10/7) (-2/5) + (3/7) (√21/5)

… = (4√10 + 3√21)/35

(which makes the last sentence in the question kind of confusing, because this expression doesn't get much simpler and it's certainly not a rational number)

i need the table so i can see

step-by-step explanation:

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)