All categories

2 years ago

Help please answer question

Mathematics

2 answers:

a_sh-v [17]2 years ago

6 0

Answer:

Step-by-step explanation:

kiruha [24]2 years ago

4 0

5 units

Hope this helps!

You might be interested in

Mike and Kim invest $12,000 in equipment to print yearbooks for schools. Each yearbook costs $5 to print and sells for $15. How

UkoKoshka [18]

<span>They must sell 1200 yearbooks to break even hope this helped</span>

5 0

2 years ago

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

Steven made $25 mowing lawns. With this money, Steven wants to save $10 and buy a new t-shirt which costs $10.81. With the rest

strojnjashka [21]

he can buy 1.52 packages.

5 0

2 years ago

Your Co. collects 50% of its sales in the month of the sales, 30% of the follow month, and 20% the second month after the sale.

Crank

Answer:

March 52
April 41

Step-by-step explanation:

In March, Your Co. will collect 20% of January's sales, 30% of February's sales, and 50% of March's sales:

.20×50 +.30×40 +.50×60 = 10 +12 +30 = 52

Similarly, in April, collections will be ...

.20×40 + .30×60 + .50×30 = 8 +18 +15 = 41

3 0

2 years ago

Read 2 more answers

What's the answers to Part A and Part B?

Westkost [7]

Answer:

there's nothing there

Step-by-step explanation:

4 0

2 years ago