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LUCKY_DIMON [66]
3 years ago
14

What is the volume of this container?

Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0

Answer:

The volume of the container is 56\ in^{3}

Step-by-step explanation:

we know that

The volume of the figure is equal to

V=Bh

where

B is the area of the base

h is the height of the figure

Let

h=4\ in

Find the area of the base B

B=(4*4)-(1*2)=14\ in^{2}

substitute

V=14*4=56\ in^{3}


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Therefore, the answer for the first box is 14/100

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If the population of a city is 8,839,469 then declines by 4% every year, what is the population in 3 years?
quester [9]
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I am in need of some help :(It looks so simple yet I don't get how to do it. Can anyone help explain?(picture of question attach
Firdavs [7]

x = 14

Explanation:

height = x

base 1 = 98

base 2 = 2

Using altitude formula in geometric mean:

base 1/height = height/base 2

98/x = x/2

\begin{gathered} \frac{98}{x}=\frac{x}{2} \\ \text{cross multiply:} \\ 2(98)\text{ = x}\times x \end{gathered}\begin{gathered} 196=x^2 \\ \text{square root both sides:} \\ \sqrt[]{196}\text{ = }\sqrt[]{x^2} \\ x\text{ = }\sqrt[]{196} \\ x\text{ = 14} \end{gathered}

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1 year ago
If a snowball melts so that its surface area decreases at a rate of 9 cm2/min, find the rate (in cm/min) at which the diameter d
zhuklara [117]

Answer:

The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.

Step-by-step explanation:

Surface area of an snowball:

An snowball has spherical format. The surface area of an sphere is given by:

S = d^2\pi

In which d is the diameter of the sphere.

In this question:

We need to differentiate S implicitly in function of time. So

\frac{dS}{dt} = 2d\pi\frac{dd}{dt}

Surface area decreases at a rate of 9 cm2/min

This means that \frac{dS}{dt} = -9

At which the diameter decreases when the diameter is 10 cm?

This is \frac{dd}{dt} when d = 10. So

\frac{dS}{dt} = 2d\pi\frac{dd}{dt}

-9 = 2(10)\pi\frac{dd}{dt}

\frac{dd}{dt} = -\frac{9}{20\pi}

\frac{dd}{dt} = -0.143

Area in cm², so diameter in cm.

The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.

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