Ask question

Ask question

All categories

3 years ago

11

Four teachers bought 10 origami books and 100 packs of oragami paper for their classrooms. They will share the cost of the items

equally. How much should each teacher pay?

1 answer:

Ksenya-84 [330]3 years ago

3 0

They will take the total price and divide it by four. So x will be the price, so the equation will be x/4=total price per person.

You might be interested in

The table below shows the percentage of households that own a microwave in different countries. Construct a stem-and-leaf plot o

S_A_V [24]

Ya then 8 |8 8
then 6 |4 9

5 0

3 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

Graph a line with a slope of.....

photoshop1234 [79]

Answer:

use google graph:)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Find the edge length of the cube Volume = 0.064 m^3<br><br>

Eva8 [605]

The edge length of the cube is 0.4

3 0

3 years ago

Read 2 more answers

Choose the function whose graph is given by:

wolverine [178]

<em>Greetings from Brasil...</em>

In a trigonometric function

F(X) = ±UD ± A.COS(Px + LR)

UD - move the graph to Up or Down (+ = up | - = down)

A - amplitude

P - period (period = 2π/P)

LR - move the graph to Left or Right (+ = left | - = right)

So:

A) F(X) = COS(X + 1)

standard cosine graph with 1 unit shift to the left

B) F(X) = COS(X) - 1 = -1 + COS(X)

standard cosine graph with 1 unit down

C) F(X) = COS(X - 1)

standard cosine graph with shift 1 unit to the right

D) F(X) = SEN(X - 1)

standard Sine graph with shift 1 unit to the right

Observing the graph we notice the sine function shifted 1 unit to the right, then

<h3>Option D</h3>

<em>(cosine star the curve in X and Y = zero. sine start the curve in Y = 1)</em>

7 0

3 years ago

Read 2 more answers