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xeze [42]
3 years ago
8

I need to know how to do this I forgot please help.

Mathematics
1 answer:
Damm [24]3 years ago
4 0

Answer:

20 units

Step-by-step explanation:

calculating volume of a rectangular prism is done by multiplying length x width x height. the prism is 5 units tall, 2 units wide, and 2 units long (or deep), so you multiply 5x2x2 to get twenty

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andriy [413]

Answer: The measure of the complement of angle x=90-x=90-30=60º

Step-by-step explanation:

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3 years ago
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Solve by Factoring<br><br> 3b^4−6b^2+27b
Ganezh [65]

Answer:

Step-by-step explanation:

hello :

3b^4−6b^2+27b = 3b(b^3-2b+9)

8 0
3 years ago
Solve the equation v+5/-9 - 4 = -6/8
tamaranim1 [39]
First you simplify both sides of the equation to make v+-41/9=-3/4.

Then add -41/9 to both sides to get the answer which is 137/36

v=137/36
4 0
3 years ago
SOMEONE PLEASE HELP ME ASAP PLEASE!!!​
marusya05 [52]

Answer:

5/6 and 6/7

Step-by-step explanation:

In the top of the fraction its going up by one each time (1,2,3,4 so the 5 and 6 would be the top half of the fraction), and in the bottom half its also going up by one but it started at 2 so its one more than the top half  (2,3,4,5, so the 6 and 7 would be the bottom half of the fraction)

4 0
3 years ago
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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
goldfiish [28.3K]

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

and so (b) the amount of sugar in the tank at time t is

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
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