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Rina8888 [55]
3 years ago
7

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 7 cubic feet

per minute. if the pool has radius 7 feet and height 12 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 8 feet
Mathematics
1 answer:
inn [45]3 years ago
8 0
7x7x12x8
=4704 yeah is that help me i don't know if is that
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The range, a measure of variability, Group of answer choices is the difference between the largest (L) and smallest (S) value in
Vlad [161]

<u>The correct answer is </u><u>all of the above.</u>

Why is the range the most convenient measure of variability?

  • Your data's spread from the lowest to the greatest value in the distribution is indicated by the range.
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  • Simply subtract the lowest value from the highest value in the data set to determine the range.

How do you find the range?

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4 0
2 years ago
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olga55 [171]

Answer:

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Step-by-step explanation:

4 0
3 years ago
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Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

8 0
3 years ago
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Likurg_2 [28]
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