3/8, -2 , 34% , -4/9 from least to greatest

Help will mark brainliest.......................................

tekilochka [14]

I'm not gonna give the answer because you have to solve it. Sorry. But I'll help you get it.

Step 1: solve the equation for each angle
Step 2: Add the totals from each angle
Step 3: Divide the total by 360
Step 4: You got your answer

I hope this helped! I'm sorry I answered really late.

3 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

marty has 1 1/3 jugs of apple juice. he uses 3/4 of the juice to make freeze pops. what fraction amount of a jus doe marty have?

Vedmedyk [2.9K]

Answer:

=2 $\frac{1}{12}$

Step-by-step explanation:

We have to make the fractions equal so we can add them

1 $\frac{1}{3}$ ×4= 1 $\frac{4}{12}$

$\frac{3}{4}$ ×3= $\frac{9}{12}$

1 $\frac{4}{12}$ + $\frac{9}{12}$ =2 $\frac{1}{12}$

5 0

2 years ago

Read 2 more answers

Solve for t <br><br> t/5/12=-3/10

natka813 [3]

T = -18 hope this helped

6 0

2 years ago

WILL GIVE BRAINLIEST! Match the equations that are the same

wolverine [178]

Answer:

1. 1/x = 8

Answer = 8x-1

2. 8x+1=3

Answer ; x=1/4

3. 7= 14/X

Answer ; x = 2

4.1/2x^2 = 2

Answer ;x=2

Step-by-step explanation:

$\frac{1}{x} =8 \\8x = 1\\\\\\8x+1=3\\Collect -like- terms \\8x =3-1\\8x = 2\\Divide -both -sides- by ;8\\\frac{8x}{8} =\frac{2}{8} \\x = 1/4\\\\\\7= \frac{14}{x} \\Cross -multiply\\7x =14\\Divide-both-sides-by-7\\x = 2\\\\\\\frac{1}{2} x^{2} =2\\\frac{x^{2} }{2} =2\\Cross-multiply\\x^{2} =4\\Squre-root -both-sides\\\sqrt{x^2}=\sqrt{4} \\x = 2\\$

6 0

3 years ago

Read 2 more answers