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Mila [183]
3 years ago
7

What is the cube root of -1000p^12q^3

Mathematics
2 answers:
Umnica [9.8K]3 years ago
7 0
∛(<span>-1000p^12q^3) 
= -10p^4 q

Hope it helps</span>
iragen [17]3 years ago
7 0

Answer:

-10p^4q

Step-by-step explanation:

Using the exponent rules:

\sqrt[n]{a^n} = a

\sqrt[3]{-1} = -1

To find the cube root of :

-1000p^{12}q^3

then;

\sqrt[3]{-1000p^{12}q^3}

We can write :

1000 = 10 \cdot 10 \cdot 10 = 10^3

p^{12} = (p^4)^3

then;

\sqrt[3]{-10^3 \cdot (p^4)^3 \cdot q^3}

Apply the exponent rules:

-10 \cdot p^4 \cdot q

⇒-10p^4q

Therefore, the cube root of -1000p^{12}q^3 is, -10p^4q

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Answer: A 4 side dice.

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Benjamin has 4 possible homerooms with the same probability. This means that every homeroom has a 1/4 = 0.25 (or 25%) probability of getting chosen.

You could emulate this situation with a 4 sided dice (are like little pyramids) where each side of the dice would represent the selection of one of the homerooms.

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3 years ago
An office building has 23 offices. The floor area of each office is 305 square feet. Which shows the equation you can use to fin
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3 0
3 years ago
Describe the characteristics of each set of numbers that make up the set of integers
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4 0
3 years ago
A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
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Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

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S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0
4 years ago
I am so confused right now plz help me
Maksim231197 [3]
You have to remember P.E.M.D.A.S. when doing problems like this.
Drew used PEMDAS so we know he isn't wrong. However, Sam didn't use PEMDAS. She multiplied 2&5 instead of raising 5 to the second. 
The correct answer for Sams' problem would be 30.
Hope this helps! :)
3 0
3 years ago
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