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k0ka [10]
3 years ago
11

Copy and compltehttps://homework.russianschool.com/resource?key=0029a

Mathematics
2 answers:
Anna35 [415]3 years ago
7 0

Answer:

A. Increases by 3000%

B. Decreases by 3000%

Step-by-step explanation:

I’m not exactly sure, but yeah.

3000% = 30

So for A. Add 3000% or 30 to it to get 70

And the same for B. 70 - 3000% or 30 to get 40 again.

I hope this helps! Good luck :)

Romashka-Z-Leto [24]3 years ago
7 0
I can’t copy so sorry, hope you have a amazing week tho! Stay safe.
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Mice21 [21]

Answer:

I say a

Step-by-step explanation:

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4 years ago
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A tank holds 300 gallons of water and 100 pounds of salt. A saline solution with concentration 1 lb salt/gal is added at a rate
Shkiper50 [21]

The amount of salt in the tank changes with rate according to

Q'(t)=\left(1\dfrac{\rm lb}{\rm gal}\right)\left(4\dfrac{\rm gal}{\rm min}\right)-\left(\dfrac{Q(t)}{300+(4-1)t}\dfrac{\rm lb}{\rm gal}\right)\left(1\dfrac{\rm gal}{\rm min}\right)

\implies Q'+\dfrac Q{300+3t}=4

which is a linear ODE in Q(t). Multiplying both sides by (300+3t)^{1/3} gives

(300+3t)^{1/3}Q'+(300+3t)^{-2/3}Q=4(300+3t)^{1/3}

so that the left side condenses into the derivative of a product,

\big((300+3t)^{1/3}Q\big)'=4(300+3t)^{1/3}

Integrate both sides and solve for Q(t) to get

(300+3t)^{1/3}Q=(300+3t)^{4/3}+C

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Given that Q(0)=100, we find

100=300+C\cdot300^{-1/3}\implies C=-200\cdot300^{1/3}

and we get the particular solution

Q(t)=300+3t-200\cdot300^{1/3}(300+3t)^{-1/3}

\boxed{Q(t)=300+3t-2\cdot100^{4/3}(100+t)^{-1/3}}

5 0
3 years ago
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lord [1]

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Step-by-step explanation:

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4 0
3 years ago
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Answer:

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Step-by-step explanation:

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4x^2 + 32x + 64=0

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6 0
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