Question

Which ratio represents the area of the smaller rectangle compared to the area of the larger rectangle? (Figure not drawn to scale)

A rectangle with the dimensions x by x plus two is inside a larger rectangle with the dimenstions two x by x squared plus six x plus eight.

x over x plus four

x plus two all over x squared plus six x plus eight

one over two times the sum of x and four

two x times the sum of x and four

Answer 1

The answer is 1/(2(x+4)) I'll type out how I got that

Answer 2

First you multiply the lengths of each rectangle to get the areas

x(x+2)

2x(x^2+6x+8)

Then to take the ratio, you divide the two:

[x(x+2)]/[2x(x^2+6x+8)]

Answer 3

It reduces to (x+2)/[2(x^2+6x+8)]

Then you factor the bottom to give:

(2+x)/[2(x+4)(x+2)]

The top cancels out with one of the bottom terms:

1/[2(x+4)]

Answer 4

Small rectangle x , x +2 dimensions
Area Smaller rectangle= x(x+2)
Large rectangle 2x , x^2 +6x +8 dimensions
Area of larger rectangle = 2x *(x^2+6x+8)
Factors of x^2 +6x +8 =(x+4)(x+2)
Ratio of smaller to larger area = [x(x+2)]/ [ 2x*(x+2)(x+4)]
1: 2(x+4)
One over two times the sum of x and 4

Answer 5

what if it was like x(x+3) 3x(x^2+7x+12)
@bme89
@rital1975
@missMob

Answer 6

@Bcoxy8

first expand \[x ^{2} + 7x +12\] into \[(x+3)(x+4)\] That will give you the ratio of \[\frac{ x(x+3) }{3x(x+4)(x+3)} \] then you can eliminate the x and (x+3) to give you...\[\frac{ 1 }{ 3(x+4) }\]