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Aleonysh [2.5K]
3 years ago
5

How to solve number 1

Mathematics
1 answer:
Mkey [24]3 years ago
4 0
You obviously solve whatever the problem is
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Algebra isn’t my best subject, please help
kogti [31]

The Distributive property states that when we have the statement a(b + c), we can "distribute" the a by multiplying a times both terms that are inside the parenthses.

So in this problem, 2(5 + 7) can be simplified by distributing the

2 through both terms that are inside the set of parenthses.

So we have (2 · 5) + (2 · 7).

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What is the area of the sector with a central angle of 97 degrees and a diameter of 10cm
vampirchik [111]

Answer:

21. 162 (rounded to nearest thousandth)

Step-by-step explanation:

Area of a sector: (degree/360) (pi*radius^2)

degree given= 97

radius= diameter/2 = 5

(97/360) (pi*5^2)

(97/360) (pi*25) = 21.16211718

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Identify the domain of the function shown in the graph.
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C. x is all real numbers

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3 years ago
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Which expression represent a number that is one fourth as great as 10-2
Nataliya [291]
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3 years ago
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an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

\dfrac{h}{r}= \dfrac{20}{8}    ( similar triangle property)

\dfrac{h}{r}= \dfrac{5}{2}

\dfrac{h}{r}= 2.5

h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

Since the water is drained  through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

4 0
3 years ago
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