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3 years ago

HELP ASAP PLS AHHHHHH

Mathematics

1 answer:

Arada [10]3 years ago

3 0

Answer:

23 tables were used that day at lunch.

Step-by-step explanation:

203 - 19 = 184

184 ÷ 8 = 23

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performance determine the slope and the y-intercept given the following linear functions. 1.f(x)=3x-10

Monica [59]

Answer:

Slope: 3

Y-Intercept: -10

Step-by-step explanation:

This equation is in slope-intercept form, it is written as y = mx + b where m is the slope and b is the y-intercept. This means 3 is the slope and -10 is the y-intercept.

3 0

3 years ago

What is an equation of the line that passes through the points (8, 2)(8,2) and (-4, 5)(−4,5)?

SOVA2 [1]

Answer:

y = -1/4x + 4

Step-by-step explanation:

(8,2) and (-4, 5)

Slope = (5 - 2)/(-4 - 8)

= 3/-12 = -1/4

Point (8,2)

b = 2 - (- 1/4)(8)

b 2+ 2 = 4

5 0

3 years ago

Use angle relationships to travel through the maze following the prompts below.

babunello [35]

Answer:

Step-by-step explanation: Not all boxes are used in the maze to prevent students from just guessing the correct ... study Use the links below to assist you in studying Parallel Lines Review Angle ... Students will be solving 16 problems on angle relationships complementary ... Angles worksheet practice questions and answers. dynamic geometry ...

4 0

3 years ago

Read 2 more answers

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

Hey, would anyone know the answers/correct matchup to this? (USA test prep)

Komok [63]

Answer:

there you go

Step-by-step explanation:

8 0

3 years ago