X+4 divided by x^2+16
why can't it be simplified?

Let's try to simplify x^2 + 16. It's a sum of two squares:

x^2 + 16 = 0

x^2 = -16

The problem is, we can't take a square root of a negative. This is where imaginary numbers come in.

Remember that square roots have a plus or minus symbol outside:

±√-16 = ±4i

Our two roots are 4i and -4i. Therefore, the trinomial simplifies to:

(x + 4i)(x - 4i)

If we attempt to divide x + 4 by these two binomials, we will find that 4 and 4i are not like terms. Therefore, we can't simplify this expression.

The volume is 56.8 triangular the

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

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

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

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

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)