Question

find the area of a rectangle with a length of 2x+2/x-4 and a width of 3x-12/x^2+6+5

Answer 1

\[\frac{ 2x+2 }{ x-4 } \] \[\frac{ 3x-12 }{ x^2+6x+5 }\]

Answer 2

A. \[\frac{ 6 }{ x+5 }\]
B\[6x+5\]
C. \[\frac{ 2x+6 }{ x-6 }\]
D. 12

Answer 3

The area of a rectangle is length times width, so with that width and length the area is:
\[\frac{2x+2}{x-4}*\frac{3x-12}{x^2+6x+5}\]
Now you need to simplify that. The best way of doing that is to start factoring stuff in the numerator and denominator and see if anything cancels.

Answer 4

im still confused :P

Answer 5

About how I got the area, or about how to simplify?

Answer 6

simplifying it

Answer 7

The area of a rectangle is length times width, so with that width and length the area is:
2x+2x−4∗3x−12x2+6x+5
Now you need to simplify that. The best way of doing that is to start factoring stuff in the numerator and denominator and see if anything cancels.

Answer 8

The area of a rectangle is length times width, so with that width and length the area is:
2x+2x−4∗3x−12x2+6x+5
Now you need to simplify that. The best way of doing that is to start factoring stuff in the numerator and denominator and see if anything cancels.

Answer 9

About how I got the area, or about how to simplify?

Answer 10

Okay, the trick is to look for anything that 'goes into' any of the terms evenly. So for example, in \(2x+2\) there is a 2 we can factor out like this \(2(x+1)\). For \(3x-12\), 3 'goes into' or divides both 3x and -12, so factored it's just \(3(x-4)\). If we put this in we get:
\[\frac{2x+2}{x-4}*\frac{3x-12}{x^2+6x+5}=\frac{2(x+1)*3(x-4)}{(x-4)*(x^2+6x+5)}\]
Now we see that \(x-4\) is both a factor in the numerator and denominator, so we cancel it, also 2*3=6 so we have
\[=6\frac{x+1}{x^2+6x+5}\]
Now the last step is to factor that second degree denominator, to do that you must find it's roots, which you know how to do. This is the process for all these kinds of problems, factor and cancel until it's impossible to simplify anymore.

Answer 11

Okay, the trick is to look for anything that 'goes into' any of the terms evenly. So for example, in 2x+2 there is a 2 we can factor out like this 2(x+1). For 3x−12, 3 'goes into' or divides both 3x and -12, so factored it's just 3(x−4). If we put this in we get:
2x+2x−4∗3x−12x2+6x+5=2(x+1)∗3(x−4)(x−4)∗(x2+6x+5)
Now we see that x−4 is both a factor in the numerator and denominator, so we cancel it, also 2*3=6 so we have
=6x+1x2+6x+5
Now the last step is to factor that second degree denominator, to do that you must find it's roots, which you know how to do. This is the process for all these kinds of problems, factor and cancel until it's impossible to simplify anymore.
a few moments ago

Answer 12

\[(2x+2)/(x-4) * (3x-12)/(x ^{2}+6x+5)\]
\[2(x+1)/(x-4) * 3(x-4)/(x+1)(x+5)\]

Answer 13

(2x+2)/(x−4)∗(3x−12)/(x2+6x+5)
2(x+1)/(x−4)∗3(x−4)/(x+1)(x+5)

Answer 14

(x+1) and (x-4) can be cancelled out..

Answer 15

(x+1) and (x-4) can be cancelled out..

Answer 16

so you will get 2*3 / (x+5)
= 6/(x+5)