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Natali5045456 [20]

3 years ago

10

Help please!!! #4 What month willl Sam have read the same amount of books as Janet

1 answer:

Vladimir [108]3 years ago

6 0

He will read books in January

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Hi someone can help me with, 5 problems of polinomial functions with degree three or more, that apply to real life

dusya [7]

Answer:8

Step-by-step explanation:

you just add the numbers up and then that si how u get your answer for your problem u are trying to work out.

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4 years ago

Do you know the answer to this

ludmilkaskok [199]

Answer:

$4x^2y\sqrt[4]{3y}$

Step-by-step explanation:

$\sqrt[4]{768x^8y^5} =$

$= \sqrt[4]{256 \times 3(x^2)^4y^4y}$

$= \sqrt[4]{4^4 \times 3(x^2)^4y^4y}$

$= 4x^2y\sqrt[4]{3y}$

8 0

3 years ago

Nichole spins each spinner below one time. 86277 What is the probability that the first spinner lands on a multiple of 3 and the

sesenic [268]

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what do you mean by this it don't make sense to me at all

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3 years ago

Name a line segment that is perpendicular to Line AC

Elanso [62]

B. should be the correct answer.

5 0

4 years ago

Read 2 more answers

Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the

ArbitrLikvidat [17]

Answer:

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let $r$ be the radius, $d$ the diameter, and $V$ the volume of the spherical balloon.

We know $\frac{dV}{dt}=20 \:{in^3/s}$ and we want to find $\frac{dr}{dt}$

The volume of a spherical balloon is given by

$V=\frac{4}{3} \pi r^3$

Taking the derivative with respect of time of both sides gives

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

We now substitute the values we know and we solve for $\frac{dr}{dt}$ :

$d=2r\\\\r=\frac{d}{2}$

$r=\frac{12}{2}=6$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044$

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and $\frac{dr}{dt}$ is:

$\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025$

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

8 0

3 years ago