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FromTheMoon [43]
3 years ago
13

3 x + 5 x + 4 - x +7 = 88

Mathematics
2 answers:
Elza [17]3 years ago
6 0
<span>3 x + 5 x + 4 - x +7 = 88

First, combine like terms.

7x + 11 = 88

Subtract 11 from both sides.

7x = 77

Divide both sides by 7.

x = 11</span>
marshall27 [118]3 years ago
4 0
3x+5x+4-x+7=88
Combining like terms
3x+5x-x+4+7=88
8x-x+11=88
7x+11=88
Subtract 11 to each side
7x+11-11=88-11
7x=77
Divided 7 to each side
7x/7=77/7
x=11
Check:
3x+5x+4-x+7=88
Substitute x with 11
3(11)+5(11)+4-11+7=88
33+55+4-11+7=88
88+4-11+7=88
92-11+7=88
81+7=88
88=88. As a result, x=11. Hope it help!
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faltersainse [42]

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Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

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\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

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