Compare: 10 quarts and 2.5 gallons *

Factor -8m^3n^3+2m^5+2m^3

Arada [10]

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2

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8

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2

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7 0

2 years ago

R(x) = 2√x s(x) = √x (rs)(4)

kati45 [8]

Answer:

8

Step-by-step explanation:

2√x • √x = 2x

2•4 = 8

5 0

2 years ago

Here are two steps from the derivation of the quadratic formula.

Lana71 [14]

What took place between the first step and the second step? - A. Taking the square root of both sides.

As we can see from the first to second step, the left side of the equation goes from being squared to not being squared. While on the right side of the equation, the square root is now being taken.

Hope this helps!! :)

6 0

3 years ago

Read 2 more answers

The diameter and height of a right circular cylinder are found at a certain instant to be 10 centimeters and 20 centimeters, res

Andrej [43]

Answer:

Step-by-step explanation:

Here's the formula for the volume of a right circular cylinder:

$V=\pi r^2h$

Here's what we are given and what we need to find:

Given that d = 10 cm, h = 20 cm, dd/dt = 1 cm/sec

Need to find dh/dt when V is constant

Since our formula has a radius in it and not a diameter but the info given is a diameter, we can use the substitution that

$2r =d$ so

$r=\frac{1}{2}d$

Now we can rewrite the formula in terms of diameter:

$V=\pi (\frac{1}{2}d)^2h$ which simplifies down to

$V=\frac{\pi }{4}d^2h$

Now we will take the derivative of this equation with respect to time using the product rule. That derivative is

$\frac{dV}{dt}=\frac{\pi }{4}[d^2*\frac{dh}{dt}+2d\frac{dd}{dt}*h]$

Now we can fill in our values. Keep in mind that if the volume is constant, there is no change in the volume, so dV/dt = 0.

$0=\frac{\pi }{4}[(100\frac{dh}{dt}+2(10)(1)(20)]$ and

$0=\frac{\pi }{4}(100\frac{dh}{dt}+400)$

Multiply both sides by pi/4 to get

$0=100\frac{dh}{dt}+400$ and solve for dh/dt:

$-4=\frac{dh}{dt}$

Interpreted within the context of our problem, this means that the volume will be constant at those given values of diameter and height when the liquid in the cylinder is dropping at a rate of 4 cm/sec.

4 0

3 years ago

Someone Help I’m in 6th grade

ad-work [718]

Answer:

2120

Step-by-step explanation:

8 0

3 years ago