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balandron [24]
3 years ago
10

The volume V (in cubic feet) of a right cylinder with a height of 3 feet and radius r (in feet) is given by V=3πr2. Solve the fo

rmula for r. Then find the radius of the cylinder when the volume is 236 cubic feet. Round your answer to the whole number.
The formula solved for r is r=
When the volume is 236 cubic feet, the radius is about __ feet.
Mathematics
1 answer:
makkiz [27]3 years ago
6 0

Answer:

118. The radius is 118 feet.

Step-by-step explanation:

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Maricel is programming an archery component of a new video game. In her code, she has created an "auto aim" feature that helps p
Nana76 [90]

Maricel's code will adjust the shot by 12º

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3 years ago
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At the beginning of a basketball season, the panthers won 20 games out of 50 games. At this rate, how many games will they win i
Anton [14]

Answer:

They will win 48 games in a 120 game season.

Step-by-step explanation:

The team won 20 games out of 50; that is a 20:50 ratio, that is equal to x:120.  x is the amount of games won in the 120 game season.

Set up the ratios like this:

\frac{20}{50} = \frac{x}{120}

cross multiply the numerators and denominators to get this:

20*120 = 50x

solve for x:

2,400 = 50x

x = 48

8 0
3 years ago
The height of a football is given by the function h of t equals negative16 t squared plus 56 t plus 3, where h of t is the heigh
kykrilka [37]

Answer:

27 feet.

Step-by-step explanation:

Graph the quadratic equation, then find the point on the graph which aligns with x=3. Y would equal 27.

8 0
2 years ago
Can I get step by step explanation on these please
OLga [1]

Aloha, thanks for using Brainly.

Step-by-step explanation:

For 1. 7/8 is 90 percent so .90

For 2+. Add the bottom together, whatever you get divide top by bottom.

The others you just divide them by each other.

6 0
3 years ago
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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