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3 years ago

If there are 16 ounces in a pound and a baby weights 10 pounds,how many ounces d9se he weight?

Mathematics

2 answers:

konstantin123 [22]3 years ago

7 0

He weighs 160 ounces. Anything multiplied by ten you can just add a zero at the end of the original number

mariarad [96]3 years ago

6 0

160 ounces because 16 (number of ounces in a pounds) multiplied by 10 (baby's weight in pounds), the answer is 160 ounces.

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5.8m

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3 years ago

What is the measure of the missing angle in this triangle<br> Yes

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If u= <4,5> and v= <0,-1>, find 2v+5u

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Answer:

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2 years ago

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a web music store offers two versions of a popular song. the size of the standard version is 2.9 megabytes (MB). the size of the

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280 downloads

Step-by-step explanation:

-let x denote the number of standard version downloads.

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-This can then be expressed as:

$x(2.9)+4x(4.8)=6188\\\\2.9x+19.2x=6188\\\\22.10x=6188\\\\x=280$

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3 years ago

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; $\mathtt{\dfrac{dh}{dt} = \dfrac{-25}{100 \pi} \ \ ft/s}$

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3 years ago