PLZZ HELP !! Select the correct answer. What are the zeros of the function y= (x-4)(x^2 - 12x+36) A. -6,4 and 6
B. 4 and 6
C. -6 and -4
D. 0, 4, and 6

The zeroes of any function are where the function touches or passes through the x-axis.

To find these zeroes, we can factor, and set each part equal to 0.

y = (x - 4)(x^2 - 12x + 36)

It looks like some of the factoring was already done for us! All we have to do is factor the quadratic. To do so, we need two numbers that multiply to equal (a * c) and add up to b. In this case those two numbers are -6 and -6.

y = (x - 4)(x - 6)(x - 6)

Finally, set each part equal to 0!

x - 4 = 0

x = 4

x - 6 = 0

x = 6

x - 6 = 0

x = 6

Therefore, the zeroes of the function are 4 and 6, option B.

Hope this helps!! :)

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)