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Bess [88]
3 years ago
5

What is the gcf of 96^5 and64^2

Mathematics
1 answer:
Arada [10]3 years ago
8 0
8 is the gcf .........
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Please answer correctly !!!! Will mark Brianliest !!!!!!!!!!!!!
NARA [144]

Answer:

2261,9 cm³

Step-by-step explanation:

First, you need to find the area of the base:

A=\pi r^2

A=36\pi cm²

Now let's multiply it for the height:

36π cm² × 20 cm = 720π cm³

= 2261,9 cm³

5 0
3 years ago
What is 58 ÷ 4083.2​
Anna007 [38]

Answer:

4083.2

Step-by-step explanation:

8 0
3 years ago
Find the area. Simplify your answer.
love history [14]

Answer:

10x + 9

Step-by-step explanation:

6 0
3 years ago
12. divied 587. hoq do i do it
Romashka [77]
For 12 divided by 587, I got 0.02044293.
For 587 divided by 12, I got 48.9166666666666.
I hope this helps.
6 0
3 years ago
Read 2 more answers
A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.
Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

t=\sqrt{\frac{2h}{a}} ,where h=height of release

a=acceleration

a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}

Where I=moment of inertia

a for hoop

a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}

a=\frac{g\sin \theta }{2}

a for Uniform solid cylinder

a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}

a=\frac{2g\sin \theta }{3}

a for spherical shell

a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}

a=\frac{3g\sin \theta }{5}

a for Uniform Solid

a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}

a=\frac{5g\sin \theta }{7}

time taken will be inversely proportional to the square root of acceleration

t_1=k\sqrt{2}=1.414k

t_2=k\sqrt{\frac{3}{2}}=1.224k

t_3=k\sqrt{\frac{5}{3}}=1.2909k

t_4=k\sqrt{\frac{7}{5}}=1.183k

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0
3 years ago
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