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soldi70 [24.7K]
3 years ago
7

Can some one please help me figure out how to calculate this styrofoam...

Mathematics
1 answer:
schepotkina [342]3 years ago
5 0

You must find the area of the circle

A = π r^2

The radius is half of the diameter, so divide 26 by 2. This equates to 13

A = π • 13^2

Multiply 13 times 13 (13^2) and it equals 169.

A = π • 169

A= 530.93

The answer is 530.93

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R+b=7 <br> 3.75r+2.75b=22.25
stiv31 [10]

Answer:

Step-by-step explanation:

hello :

r+b=7... (*)

3.75r+2.75b=22.25.... (**)

use (*) :  r = 7 - b

put this value in (**) :

3.75(7 - b)+2.75b =  22.25

26.25 -375b +2.75b = 22.5

-375b +2.75b = 22.5 - 26.25

- b = -3.75

so : b  = 3.75

but : r = 7 - b

r = 7 - 3.75

r = 3.25

6 0
3 years ago
Please help pleaseeee I really need answers only
stiks02 [169]

☁️ Answer ☁️

5. IS E (none of these)

Because I got:

-125y^5/162x

6. X = 5/3

Hope it helps.

Have a nice day noona/hyung.

7 0
2 years ago
Read 2 more answers
Substitution method for 7x-2y=1 &amp; x-2y=1
RUDIKE [14]
We want to use the Substitution method to solve
7x - 2y = 1          (1)
x - 2y = 1           (2)

From equation (1), obtain
2y = 7x - 1          (3)

Substitute (3) into (2).
x - (7x - 1) = 1
-6x + 1 = 1
-6x = 0
x = 0

From (3), obtain
2y = 7*0 - 1 = -1
y = -1/2

Answer: (0, -1/2)  or x=0, y= -1/2.

6 0
3 years ago
Am I dealing with harmonic stuff (trig)?
pochemuha

You just need to solve for when y=0:

\dfrac{\cos8t-9\sin8t}4=0\implies\cos8t-9\sin8t=0

\implies\cos8t=9\sin8t

\implies\dfrac19=\dfrac{\sin8t}{\cos8t}=\tan8t

\implies8t=\tan^{-1}\dfrac19+n\pi

\implies t=\dfrac18\tan^{-1}\dfrac19+\dfrac{n\pi}8

where n is any integer. We only care about when 0\le t\le1, which happens for n\in\{0,1,2\}.

t=\dfrac18\tan^{-1}\dfrac19\approx0.01

t=\dfrac18\tan^{-1}\dfrac19+\dfrac\pi8\approx0.41

t=\dfrac18\tan^{-1}\dfrac19+\dfrac\pi4\approx0.80

8 0
3 years ago
Why do i have to learn the quadratic equation?
marysya [2.9K]
<h2><u>↑I DON'T KNOW↓</u></h2>
  1. <em>IDK</em>
  2. <em>IDK</em>
  3. <em>IDK</em>
  4. <em>IDK</em>
  5. <em>IDK</em>
  6. <em>IDK</em>
  7. <em>IDK</em>
  8. <em>IDK</em>
  9. <em>IDK</em>
  10. <em>IDK</em>
5 0
3 years ago
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