If x¹=xcosA+ysinA and y¹=xsinA-ycosA, show that (x¹)²+(y¹)²=x²+y²​

Expanding each square on the left side, you have

(x cos(A) + y sin(A))² = x² cos²(A) + 2xy cos(A) sin(A) + y² sin²(A)

(x sin(A) - y cos(A))² = x² sin²(A) - 2xy sin(A) cos(A) + y² cos²(A)

so that adding them together eliminates the identical middle terms and reduces to the sum to

x² cos²(A) + y² sin²(A) + x² sin²(A) + y² cos²(A)

Collecting terms to factorize gives us

(y² + x²) sin²(A) + (x² + y²) cos²(A)

(x² + y²) (sin²(A) + cos²(A))

and sin²(A) + cos²(A) = 1 for any A, so we end up with

x² + y²

as required.

30 people weigh 160 pounds or more

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.