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lozanna [386]
3 years ago
12

How do i find the volume of a cone

Mathematics
2 answers:
ratelena [41]3 years ago
7 0
You can use the formula 3.14 times the radius squared by height over 3
enyata [817]3 years ago
3 0
V=PiR^2H/3

Pi = 3.14 (use pi symbol on calculator)
R is the radius if the cones bottom
H is the height of the cone
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Please please help thanks ❤️
Svetlanka [38]
C. 48 inches 
Add all the sides to find the perimeter
7 0
3 years ago
Please help thanks thanks thanks ty
Archy [21]

Answer:

104 degrees

Step-by-step explanation:

First let's start by finding Angle DGC. Since Angle FGD and DGC are a linear pair, Angle DGC = 180-90 = 90 degrees.

Next we need to find GDC

Angles in a triangle add up to 180 degrees therefore...

Angle GDC + DCG + DGC = 180

Plug in the values we found into the equation

Angle GDC + 37 + 90 = 180

Angle GDC + 127 = 180

Angle GDC = 53

Therefore Angle ADC = 53 + 51 = 104 degrees

Since ABCD is a parallelogram, then opposite angles are equal therefore...

Angle B = 104 degrees

3 0
3 years ago
HELP!!! I HAVE TO DO THIS!! WILL GIVE
sertanlavr [38]

Answer:

1. 9.576

2. 17.442

3. 45.29

4. 79.294

5. 30.736

6. 30.8

7. 31.5

8. 25.85

9. 53.29

10. 49.184

11. 15.18

12. 15.573

13. 17.075

14. 74.06

15. 9.01

Step-by-step explanation:

Use a multiplication method to find the answers. I would put it up here but i dont know how.

3 0
3 years ago
14/15 divided by 7/5
Charra [1.4K]

9514 1404 393

Answer:

  2/3

Step-by-step explanation:

There are a couple of different ways that division of fractions can be done.

1) "invert and multiply"

  \dfrac{14}{15}\div\dfrac{7}{5}=\dfrac{14}{15}\times\dfrac{5}{7}\\\\=\dfrac{14}{7}\times\dfrac{5}{15}=\boxed{\dfrac{2}{3}}

__

2) use the numerators when the denominators are the same

  \dfrac{14}{15}\div\dfrac{7}{5}=\dfrac{14}{15}\div\dfrac{21}{15}\\\\=\dfrac{14}{21}=\boxed{\dfrac{2}{3}}

6 0
3 years ago
Write as a percent. (if necessary, round to the nearest tenth of a percent.)13/30
labwork [276]
The correct answers is 43.33%
6 0
3 years ago
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