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2 years ago

5

Ava plays on the softball team at her school. The team won 8 of its first 12 games. At that rate, how many games will Ava's team

win if it plays 60 games in all?

1 answer:

GaryK [48]2 years ago

3 0

Answer:

She wins 40 games

Step-by-step explanation:

I'm not the best at explaining but 5x12=60 so then we multiply 8 by 5 aswell and bam we have 40.

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If the diameter of a circle with a circumference of 19.5 inches is reduced by three, what would be the circumference of the new

Elanso [62]

It is given that the circumference of the circle is 19.5 inches. Let the diameter is d inches .

And the formula of circumference is

$C= \pi d$

Substituting the value of C, we will get

$19.5= \pi d
\\
d = \frac{19.5}{ \pi}$

When the diameter increased by 3, then the circumference is

$C = \pi ( \frac{19.5}{ \pi} +3})=Approx 29 inches$

And the circumference , when diameter is increased by 3 is 29 inches .

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3 years ago

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Addy is taking a test with multiple choice and True/False questions. On the test, Addy got a total of twenty questions correct.

vovangra [49]

Answer:

12 multiple choice and 8 true/false

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2 years ago

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19 What is the product of the complex number 2 +

hjlf

Answer:

D

Step-by-step explanation:

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2 years ago

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

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

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

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

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

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2 years ago

Using the vertical line test. Is this a function?

marta [7]

Answer:

Can't get one(Read explanation)

Step-by-step explanation:

What you do to find out if its a function is see if any of the dots are in the same x value. If they are, its not a function. If they aren't it is a function.

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2 years ago