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podryga [215]
3 years ago
12

Help me!!! I am so confused

Mathematics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

\boxed{x^{2} + 2x - 8}

Step-by-step explanation:

6. Practice

The dimensions of the current park are x long and x wide.

The new park will be 4 longer and 2 thinner.

Its new dimensions will be x + 4 long and x – 2 wide.

Its new area will be

A = width × length = (x – 2)(x + 4)

Find the product

\begin{array}{lll}\textbf{Steps} & \textbf{Problem: }(x - 2)(x + 4) & \\\textbf{1. List variables} & a = x - 2 & \\ & b = x & \\ & c = 4 &\\\textbf{2. Distribute (x - 2)} & (x -2)(x + 4)\\ & = (x - 2)(x) + (x - 2)(4)\\\textbf{3. Distribute x and 4} & x^{2} -2x + 4x - 8\\\textbf{4. Combine like terms}& x^{2} + 2x - 8\\\end{array}\\\text{The area of the updated skatepark will be }\boxed{\mathbf{ x^{2} + 2x - 8}}

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N/4 &lt; -1 on a number line<br> -10x&gt;-100 on a number line <br> 5x&gt;20 on a number line
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N < -4

So, our N goes left from -4 on a number line.

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Please help with these partial fractions!!!
VARVARA [1.3K]

a. Factorize the denominator:

\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}

Then we're looking for a,b such that

\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}

\implies x+14=a(x+2)+b(x-4)

If x=4, then 18=6a\implies a=3; if x=-2, then 12=-6b\implies b=-2. So we have

\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}

as required.

b. Same setup as in (a):

\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}

We want to find a,b,c such that

\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}

Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant b' such that the numerator of the second term is b'x+b. We always want the polynomial in the numerator to have degree 1 less than the degree of the denominator. But we would end up determining b'=0 anyway.

\implies-3x^2+5x+6=ax(x+1)+b(x+1)+cx^2

If x=0, then b=6; if x=-1, then c=-2. Expanding everything on the right then gives

-3x^2+5x+6=ax^2+ax+bx+b+cx^2=(a-2)x^2+(a+6)x+6

which tells us a-2=-3 and a+6=5; in both cases, we get a=-1. Then

\dfrac{-3x^2+5x+6}{x^2(x+1)}=-\dfrac1x+\dfrac6{x^2}-\dfrac2{x+1}

as required.

5 0
4 years ago
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