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3 years ago

Which situation could be represented by the graph?

Mathematics

2 answers:

Anarel [89]3 years ago

6 0

Answer:

Step-by-step explanation:

before looking at the options, what does the graph show? To summarize, numbers between 8 and 11 including 8 but excluding 11, so keep that in mind when looking at the options.

A) this oen is 6 to 11, so we don't even need to look at if it includes or exludes end points.

B) this is between 8 and 11, but it includes both 8 and 11 as far as I can tell, so that's not right either, but closer.

C) Between 8 and 11 again, this time though it has to include 8 but can't include 11. So that looks like our answer. Just in case let's look at the last one.

D)This one is basically the opposite of what we want. 8 and under or 11 and over, so 8-11 is explicitly excluded.

So yep, C is the answer.

Citrus2011 [14]3 years ago

3 0

Answer:

the answer is c

Step-by-step explanation:

I took the test

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Answer:

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Step-by-step explanation:

You have the following curves:

$y_1=\frac{4}{9}x^2=f(x)\\\\y_2=\frac{13}{9}-x^2=g(x)$

In order to calculate the solid of revolution bounded by the previous curves and the x axis, you use the following formula:

$V=\pi \int_a^b [(g(x))^2-(f(x))^2]dx$ (1)

To determine the limits of the integral you equal both curves f=g and solve for x:

$f(x)=g(x)\\\\\frac{4}{9}x^2=\frac{13}{9}-x^2\\\\\frac{4}{9}x^2+x^2=\frac{13}{9}\\\\\frac{13}{9}x^2=\frac{13}{9}\\\\x=\pm 1$

Then, the limits are a = -1 and b = 1

You replace f(x), g(x), a and b in the equation (1):

$V=\pi \int_{-1}^{1}[(\frac{13}{9}-x^2)^2-(\frac{4}{9}x^2)^2]dx\\\\V=\pi \int_{-1}^1[\frac{169}{81}-\frac{26}{9}x^2+x^4-\frac{16}{81}x^4]dx\\\\V=\pi \int_{-1}^1 [\frac{169}{81}-\frac{26}{9}x^2+\frac{65}{81}x^4]dx\\\\V=\pi [\frac{169}{81}x-\frac{26}{27}x^3+\frac{65}{405}x^5]_{-1}^1\\\\V\approx8.06\ cubed\ units$

The volume of the solid of revolution is approximately 8.06 cubed units

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