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Nata [24]
3 years ago
9

Simplify The Expression 16(1/2x)^4

Mathematics
1 answer:
Bond [772]3 years ago
8 0
16(1/2x)^4
Answer: X^4
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Part A: For each problem, write an expression using parenthesis, and then solve.
slavikrds [6]
<span>1. $119 2. 64 beads 3. 4 candles 4</span>
6 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
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
abrl,belle and cindy have $408 altogeter.belle has $7 more than cindy and $5 more than abel.how much does abel have?
faltersainse [42]
ABC = 408 
C + 7 = B
A + 5 = B 

B - 7 + B - 5 + B = 408
3B - 12 = 408
     + 12 = +12
3B = 420
/3   = /3
B = 120
A = 115
C = 113
The answer is A = 115, Abel has $ 115

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3 years ago
Kristen worked 39 hours in 6 days
Fittoniya [83]
<h2>39/6 = 6.5</h2><h2 /><h2>6 hours 30 minutes</h2>
5 0
3 years ago
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Answer:

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