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tatyana61 [14]
3 years ago
9

Please help. 10 points and brainliest.

Mathematics
1 answer:
Norma-Jean [14]3 years ago
7 0

Answer:

D

Step-by-step explanation:

since you can't divide by zero, you get a vertical asymptote (x=something) in effect

you can't divide by x+3

if x=-3, thus answers A and B are wrong

now lets look at the coefficients of 2/(x-3)

the horizontal asymptote of -1 exists in the rational equation D, since it is transformed down one unit, but has the same degrees in 2/(x+3), specifically, 1/1=1, thus we can use y=-1

here is a graph to prove everything

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rosijanka [135]
\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx

Notice that x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1). Decompose the integrand into partial fractions:

\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}
=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}

Integrating term-by-term, you get

\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx
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5 0
3 years ago
Solve the equation <br> |x-3| -10=-5
Deffense [45]

Answer:

x=8                     x=-2

Step-by-step explanation:

|x-3| -10=-5

Add 10 to each side

|x-3| -10+10=-5+10

|x-3| =5

Now separate into two equations , one positive and one negative

x-3 = 5           x-3 = -5

Add 3 to each side

x-3+3 = 5+3    x-3+3 = -5 +3

x=8                     x=-2

5 0
3 years ago
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mr_godi [17]
-50x :)
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2 years ago
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3 years ago
Suppose that y varies inversely with x, and y = 2 when x = 4. What is an<br>variation?​
Alexeev081 [22]

Answer:

xy=8

Step-by-step explanation:

The equation for inverse variation is

xy = k where k is the constant of variation

4*2 = k

8=k

The equation is

xy=8

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3 years ago
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