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Eddi Din [679]
3 years ago
11

Functions... what is NOT true?

Mathematics
1 answer:
Lynna [10]3 years ago
8 0

The answer for what is NOT true would be..

B.  A sequence is a function whose domain is the set of real numbers.

&

C. A function is a relation in which each value of the input variable is paired with exactly one value of the output variable.


I hope this helps have a great day.

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Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
Complete the equation, by supplying the missing exponent.
Vesna [10]
The answer is 1.927x102
4 0
3 years ago
Read 2 more answers
PLEASE HELP ME I NEED HELP FAST
Alexxandr [17]
The answer would be A if i am correct can i get most brainliest

6 0
3 years ago
1. Juan bought a pair of shoes costing P923.00 and a tie costing P275.00. He gave the seller P2, 000.00. How much change did he
EleoNora [17]
2000.00-923.00-275.00=802.00
4 0
3 years ago
What is the inverse of the function below?<br><br> f(x) = 2^x + 6
Helen [10]

The inverse of this function would be f(x) = \frac{Log(x - 6)}{Log2}.



You can find the value of any inverse function by switching the f(x) and the x value. Then you can solve for the new f(x) value. The end result will be your new inverse function. The step-by-step process is below.



f(x) = 2^{x} - 6 ----> Switch f(x) and x



x = 2^{f(x)} - 6 ----> Add 6 to both sides



x + 6 = 2^{f(x)} -----> Take the logarithm of both sides in order to get the f(x) out of the exponent



Log(x + 6) = f(x)Log2 ----> Now divide both sides by Log2



\frac{Log(x + 6)}{Log2} = f(x) ----> And switch the order for formatting purposes.



f(x) = \frac{Log(x + 6)}{Log2}



And that would be your new inverse function.

6 0
3 years ago
Read 2 more answers
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