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11Alexandr11 [23.1K]
3 years ago
14

The radius of a circular puddle is growing at a rate of 25 cm/sec.

Mathematics
1 answer:
tatiyna3 years ago
8 0

Answer:

Step-by-step explanation:

a)area A=pi r^2

rate of change of area =dA/dt =2 pi r dr/dt

given dr/dt =25 ,r =50

rate of change of area =dA/dt =2 pi *50 *25 =2500pi=7854

area growing 7854 cm2/s

b)area A=pi r^2

rate of change of area =dA/dt =2 pi r dr/dt

given dr/dt =25 ,A =64

pi r^2 =64

=>r =8/√pi

rate of change of area =dA/dt =2 pi *(8/√pi) *25 =400√pi=708.98

area growing 708.98 cm2/s

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Write two equations. the first equation should express the area of the square (a as a function of the length (l of a side. the s
Sergeu [11.5K]

Answer:

a = l²

v = s³

Step-by-step explanation:

The area of a rectangle is the product of its length and width. When that rectangle is a square, the length and width are the same. Here, they are given as "l". Then the area of the square is ...

a = l·l = l²

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The volume of a cuboid is the product of its height and the area of its base. A cube of edge length s has a square base of side length s and a height of s. Then its volume will be ...

v = s·(s²) = s³

The two equations you want are ...

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6 0
3 years ago
Help! What is the average rate of change of f(x)=x^2+3x+6 over the interval -3 less-than-or-equal-to x less-than-or-equal-to 3?
Makovka662 [10]

Answer:

The average rate of change of the function over the interval is 5.

Step-by-step explanation:

Average rate of change of a function:

The average rate of change of a function f(x) over an interval [a,b] is given by:

A = \frac{f(b)-f(a)}{b-a}

Interval -3 less-than-or-equal-to x less-than-or-equal-to 3

This means that a = -3, b = 3

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So

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f(a) = f(-3) = (-3)^2 + 3(-3) + 6 = 9 - 9 + 6 = 6

Average rate of change

A = \frac{f(b)-f(a)}{b-a} = \frac{24+6}{3-(-3)} = \frac{30}{6} = 5

The average rate of change of the function over the interval is 5.

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3 years ago
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The correct answer is 42.
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I think the answer is:
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