4 4/9 simplified into the smallest from

Brennan has already run 16 miles on his own, and he expects to run 3 miles during each track practice. How many track practices

pentagon [3]

Answer:

It would take Brennan 4 track practice runs.

Step-by-step explanation:

He has already run 16 miles and needs to run 28 in total. If we are counting the 16 in the 28 miles, he would need to run only 12 more to reach his goal. And if he runs 3 miles a run, it would take him about 4 runs to finish his 28-mile goal.

(28-16=12 and 12/3=4)

8 0

3 years ago

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PLZ HELP ME Use the distribute property to subtract and combine like terms <br> (5x)-(3x² + 3x)

Montano1993 [528]

Answer:

look up the distributive property and solve the problem.

Step-by-step explanation:

7 0

3 years ago

There are two correct answers

UNO [17]

Answer:

I think d no p=4 and q=5

Step-by-step explanation:

this might help you

7 0

3 years ago

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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

The regular octagon has a perimeter of 122.4 cm. Which statements about the octagon are true? Select two options. The length of

KATRIN_1 [288]

The correct answers on edge are C and E

6 0

4 years ago

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