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3 years ago

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