First you need to get a common denominator. The first two problems already have a common denominator, so you just have to find the sum and difference of the two problems, respectively.

6.) (9y-3)/(y^2-5)

8.) 3/xy^3

Two find a common denominator, find the smallest value that has BOTH denominators as factors.

10.) 8x^3*y^3 is the common denominator

Answer: 2/(8*x^3*y^3)

12.) Factor x^2 + 4x + 4 to get (x+2)(x+2)

Therefore, x+2 is already part of x^2 + 4x + 4

(x+4)/(x^2+4x+4)

14.) Factor both denominators..

4(y+2) and y(y+2)

So the common denominator would be 4y(y+2)

The first fraction is missing a "y" so multiply the numerator by y to get y^2

The second fraction is missing a "4" so multiply the numerator by 4 to get 4

(y^2 - 4)/(4y(y+2))

So can factor the numerator to get (y+2)(y-2). There if a (y+2) on top and bottom so they cancel out.

Answer: (y-2)/4y

6 0

4 years ago

The length of a picture is 15.75 inches shorter than twice the width. If the perimeter of the picture is 106.5 inches, find its

Vladimir79 [104]

Answer:

Length = 45.37 in

Width = 30.56 in

Step-by-step explanation:

Let the width of the picture be 'w' inches.

Given:

Perimeter of the picture is, $P=106.5\ in$

Length is 15.75 inches shorter than twice the width. So, length is:

$l=2w-15.75$

Now, we know that the perimeter of a rectangular shape is given as the sum of all of its side. So, the picture is also of rectangular shape and hence the perimeter is given as:

$P=2l+2w$