Question

HELP!!!! I NEED HELP WITH THIS.

Answer 1

Answer:

$\large\boxed{A=x^2+23x+49}$

Step-by-step explanation:

Subtract the area of a square (x + 1) × (x + 1)

from the area of a rectangle (x + 10) × (2x + 5)

The area of a square:

$A_s=(x+1)(x+1)$            use FOIL: (a + b)(c + d) = ac + ad + bc + bd

$A_s=(x)(x)+(x)(1)+(1)(x)+(1)(1)=x^2+x+x+1=x^2+2x+1$

The area of a rectangle:

$A_r=(x+10)(2x+5)$              use FOIL

$A_r=(x)(2x)+(x)(5)+(10)(2x)+(10)(5)=2x^2+5x+20x+50=2x^2+25x+50$

The area of a figure:

$A=A_r-A_s$

Substitute:

$A=(2x^2+25x+50)-(x^2+2x+1)=2x^2+25x+50-x^2-2x-1$

combine like terms

$A=(2x^2-x^2)+(25x-2x)+(50-1)=x^2+23x+49$