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kozerog [31]
3 years ago
8

Find the area between the graph of the function and the x-axis over the given interval, if possible.

Mathematics
1 answer:
vodomira [7]3 years ago
7 0

Answer:

A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5

So then the integral converges and the area below the curve and the x axis would be 5.

Step-by-step explanation:

In order to calculate the area between the function and the x axis we need to solve the following integral:

A = \int_{-\infty}^1 \frac{5}{(x-2)^2}

For this case we can use the following substitution u = x-2 and we have dx = du

A = \int_{a}^b \frac{5}{u^2} du = 5\int_{a}^b u^{-2}du

And if we solve the integral we got:

A= -\frac{5}{u} \Big|_a^b

And we can rewrite the expression again in terms of x and we got:

A = -\frac{5}{x-2} \Big|_{-\infty}^1

And we can solve this using the fundamental theorem of calculus like this:

A= -\frac{5}{-1} - \lim_{x\to\infty} \frac{5}{x-2} = 5-0 = 5

So then the integral converges and the area below the curve and the x axis would be 5.

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Nadia finds out her favorite horse family population is increasing at a constant rate. The horse family was at 24 in 2011 and is
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Answer:

The equation is p = (8/3) t + 24

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Step-by-step explanation:

In 3 years  the family increased by 32 - 24 = 8.

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How do I factor out a constant before factoring out a quadratic? <br>Factor completely: 3w^2-3w-90
pantera1 [17]
3w^2-3w-90 =0 \ \ \:3 \\ \\w^2- w-30 =0 \\ \\a=1 , \ b= -1 , c= -30 \\ \\\Delta =b^2-4ac = (-1)^2 -4\cdot1\cdot (-30) =1+120=121 \\ \\w_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{1-\sqrt{121}}{2 }=\frac{ 1-11}{2}=\frac{-10}{2}=-5

w_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{1+\sqrt{121}}{2 }=\frac{ 1+11}{2}=\frac{12}{2}=6 \\ \\ Answer:\\ \\ w^2- w-30 =(x+5)(x-6)
 

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3 years ago
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