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kolbaska11 [484]
2 years ago
15

Determine the sum -46.38 (-24.6)

Mathematics
1 answer:
Zarrin [17]2 years ago
8 0
Well, this equation as a sum would be -46.38 + -24.6 
so the answer would be -70.98. 

I hope this helps, and have a fantastic day!
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ankoles [38]

Answer:

1) m=-7/5

2) m=-7/5

3) parallel

Step-by-step explanation:

the first part:

graph the points and then find the slope

m=-7/5

the second part:

slope= m in y=mx+b

m=-7/5

the last part:

they are parallel because they have the same slope

5 0
3 years ago
Read 2 more answers
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

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(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}

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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}

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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.

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3 years ago
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quester [9]
It would be 0.2222 so it's your answer
5 0
3 years ago
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castortr0y [4]

Answer:

36 cm

Step-by-step explanation:

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The following is an expression.
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