5. factor the expression. 6x2 + 5x + 1 a)(3x + 1)(2x – 1) b)(3x – 1)(2x – 1) c)(3x – 1)(2x + 1) d)(3x + 1)(2x + 1)

The factored version of the equation 6x^2 + 5x + 1 is equal to (3x+1)(2x+1).

Step-by-step explanation:

Here is the reason why that equation is correct:

Use the FOIL method for each factored equation to see if it equals 6x^2 + 5x + 1.

(3x + 1)(2x - 1)

3x * 2x = 6x^2.

3x * -1 = -3x.

1 * 2x = 2x.

1 * -1 = -1.

6x^2 - x - 1.

(3x - 1)(2x - 1)

3x * 2x = 6x^2.

3x * -1 = -3x.

-1 * 2x = -2x.

-1 * -1 = 1.

6x^2 - 5x + 1.

(3x -  1)(2x + 1)

3x * 2x = 6x^2.

3x * 1 = 3x.

-1 * 2x = -2x.

-1 * 1 = -1.

6x^2 + x - 1.

(3x + 1)(2x + 1)

3x * 2x = 6x^2.

3x * 1 = 3x.

1 * 2x = 2x.

1 * 1 = 1.

6x^2 + 5x + 1.

D contains the only factored expression that equals 6x^2 + 5x + 1 as a result using the FOIL method.

-4

step-by-step explanation:

(-4+8)=-(4-8)=-8+4=-4

< 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)