Answer:
Ok so...
Mean: 68.76
Median: 72
Range: 85
Step-by-step explanation:
Metallic ribbon: 16$ for 4ft
Dividing both sides by four, we get:
1 feet -- 4 dollars
White ribbon 3$ for 1 feet
1 feet -- 3 dollars
<h2><u><em>
Therefore, the Metallic ribbon costs more per feet.</em></u></h2>
Answer:
1.
= ![\frac{2x^2+10x+6}{(x+3)(x+5)}\\](https://tex.z-dn.net/?f=%5Cfrac%7B2x%5E2%2B10x%2B6%7D%7B%28x%2B3%29%28x%2B5%29%7D%5C%5C)
2.
= ![\frac{1}{(x+2)(x-4)}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28x%2B2%29%28x-4%29%7D)
3.
= ![\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}](https://tex.z-dn.net/?f=%5Cfrac%7B-3x%5E2-7x-4%7D%7B%28x%2B3%29%28x-3%29%28x-2%29%7D)
4.
= ![\frac{1}{(x-2)(x-4)}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28x-2%29%28x-4%29%7D)
Step-by-step explanation:
1. ![\frac{x}{x+3}+\frac{x+2}{x+5}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bx%2B3%7D%2B%5Cfrac%7Bx%2B2%7D%7Bx%2B5%7D)
Taking LCM of (x+3) and (x+5) which is: (x+3)(x+5)
![=\frac{x(x+5)+(x+2)(x+3)}{(x+3)(x+5)}\\=\frac{x^2+5x+(x)(x+3)+2(x+3)}{(x+3)(x+5)} \\=\frac{x^2+5x+x^2+3x+2x+6}{(x+3)(x+5)} \\=\frac{x^2+x^2+5x+3x+2x+6}{(x+3)(x+5)} \\=\frac{2x^2+10x+6}{(x+3)(x+5)}\\](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bx%28x%2B5%29%2B%28x%2B2%29%28x%2B3%29%7D%7B%28x%2B3%29%28x%2B5%29%7D%5C%5C%3D%5Cfrac%7Bx%5E2%2B5x%2B%28x%29%28x%2B3%29%2B2%28x%2B3%29%7D%7B%28x%2B3%29%28x%2B5%29%7D%20%5C%5C%3D%5Cfrac%7Bx%5E2%2B5x%2Bx%5E2%2B3x%2B2x%2B6%7D%7B%28x%2B3%29%28x%2B5%29%7D%20%5C%5C%3D%5Cfrac%7Bx%5E2%2Bx%5E2%2B5x%2B3x%2B2x%2B6%7D%7B%28x%2B3%29%28x%2B5%29%7D%20%5C%5C%3D%5Cfrac%7B2x%5E2%2B10x%2B6%7D%7B%28x%2B3%29%28x%2B5%29%7D%5C%5C)
Prove closure: The value of x≠-3 and x≠-5 because if there values are -3 and -5 then the denominator will be zero.
2. ![\frac{x+4}{x^2+5x+6}*\frac{x+3}{x^2-16}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B4%7D%7Bx%5E2%2B5x%2B6%7D%2A%5Cfrac%7Bx%2B3%7D%7Bx%5E2-16%7D)
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
Factors of x^2+5x+6 = x^2+3x+2x+6 = x(x+3)+2(x+3) =(x+2)(x+3)
Putting factors
![=\frac{x+4}{(x+3)(x+2)}*\frac{x+3}{(x-4)(x+4)}\\\\=\frac{1}{(x+2)(x-4)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bx%2B4%7D%7B%28x%2B3%29%28x%2B2%29%7D%2A%5Cfrac%7Bx%2B3%7D%7B%28x-4%29%28x%2B4%29%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B%28x%2B2%29%28x-4%29%7D)
Prove closure: The value of x≠-2 and x≠4 because if there values are -2 and 4 then the denominator will be zero.
3. ![\frac{2}{x^2-9}-\frac{3x}{x^2-5x+6}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7Bx%5E2-9%7D-%5Cfrac%7B3x%7D%7Bx%5E2-5x%2B6%7D)
Factors of x^2-9 = (x)^2-(3)^2 = (x-3)(x+3)
Factors of x^2-5x+6 = x^2-2x-3x+6 = x(x-2)+3(x-2) =(x-2)(x+3)
Putting factors
![\frac{2}{(x+3)(x-3)}-\frac{3x}{(x+3)(x-2)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B%28x%2B3%29%28x-3%29%7D-%5Cfrac%7B3x%7D%7B%28x%2B3%29%28x-2%29%7D)
Taking LCM of (x-3)(x+3) and (x-2)(x+3) we get (x-3)(x+3)(x-2)
![\frac{2(x-2)-3x(x+3)(x-3)}{(x+3)(x-3)(x-2)}](https://tex.z-dn.net/?f=%5Cfrac%7B2%28x-2%29-3x%28x%2B3%29%28x-3%29%7D%7B%28x%2B3%29%28x-3%29%28x-2%29%7D)
![=\frac{2(x-2)-3x(x+3)}{(x+3)(x-3)(x-2)}\\=\frac{2x-4-3x^2-9x}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-9x+2x-4}{(x+3)(x-3)(x-2)}\\=\frac{-3x^2-7x-4}{(x+3)(x-3)(x-2)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%28x-2%29-3x%28x%2B3%29%7D%7B%28x%2B3%29%28x-3%29%28x-2%29%7D%5C%5C%3D%5Cfrac%7B2x-4-3x%5E2-9x%7D%7B%28x%2B3%29%28x-3%29%28x-2%29%7D%5C%5C%3D%5Cfrac%7B-3x%5E2-9x%2B2x-4%7D%7B%28x%2B3%29%28x-3%29%28x-2%29%7D%5C%5C%3D%5Cfrac%7B-3x%5E2-7x-4%7D%7B%28x%2B3%29%28x-3%29%28x-2%29%7D)
Prove closure: The value of x≠3 and x≠-3 and x≠2 because if there values are -3,3 and 2 then the denominator will be zero.
4. ![\frac{x+4}{x^2-5x+6}\div\frac{x^2-16}{x+3}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B4%7D%7Bx%5E2-5x%2B6%7D%5Cdiv%5Cfrac%7Bx%5E2-16%7D%7Bx%2B3%7D)
Factors of x^2-5x+6 = x^2-3x-2x+6 = x(x-3)-2(x-3) = (x-2)(x-3)
Factors of x^2-16 = (x)^2 -(4)^2 = (x-4)(x+4)
![\frac{x+4}{(x-2)(x+3)}\div\frac{(x-4)(x+4)}{x+3}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B4%7D%7B%28x-2%29%28x%2B3%29%7D%5Cdiv%5Cfrac%7B%28x-4%29%28x%2B4%29%7D%7Bx%2B3%7D)
Converting ÷ sign into multiplication we will take reciprocal of the second term
![=\frac{x+4}{(x-2)(x+3)}*\frac{x+3}{(x-4)(x+4)}\\=\frac{1}{(x-2)(x-4)}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bx%2B4%7D%7B%28x-2%29%28x%2B3%29%7D%2A%5Cfrac%7Bx%2B3%7D%7B%28x-4%29%28x%2B4%29%7D%5C%5C%3D%5Cfrac%7B1%7D%7B%28x-2%29%28x-4%29%7D)
Prove Closure: The value of x≠2 and x≠4 because if there values are 2 and 4 then the denominator will be zero.