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lesya [120]
3 years ago
9

Can someone please take me step by step to find x?

Mathematics
2 answers:
Aleksandr [31]3 years ago
8 0
-(7x + 4)- 2x + 2 = 52

Distribute the negative sign:
-7x -4 - 2x + 2 = 52

Combine like terms:
-9x - 2 = 52

Add 2 to both sides:
-9x = 54

Divide by -9:
x = -6

Answer: x = -6
klemol [59]3 years ago
7 0
1. distribute the negative sign it become -7x - 4
2. combine like term
3. you would know what to do from there ;)
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11. Charles's Law states that at constant pressure, the volume of a fixed amount of gas varies
Zanzabum

Answer:

a.V=5/6T

b. 350 ml

Step-by-step explanation:

a.

V∝T

or V=kT

where k is a constant.

V=250 ml

T=300° K

250=300 k

k=250/300=5/6

so V=5/6 T

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6 0
3 years ago
Find the solution to the differential equation<br><br> dB/dt+4B=20<br><br> with B(1)=30
natita [175]

Answer:

The solution of the differential equation is B=5+25e^{-4t+4}

Step-by-step explanation:

The differential equation \frac{dB}{dt}+4B=20 is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

y'(t)=g(t)\cdot h(y)

where <em>g(t)</em> and <em>h(y) </em>are given functions<em>. </em>

We can rewrite our differential equation in the form of a first-order separable ODE in this way:

\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4

Integrating both sides

\frac{1}{5-B}\frac{dB}{dt}=4\\\frac{1}{5-B}\cdot dB=4\cdot dt\\\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {4} \, dt

The integral of left-side is:

\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C

The integral of right-side is:

\int\limits {4} \, dt = 4t + C

We can join the constants, and this is the implicit general solution

-\ln \left|5-B\right|+C=4t + C\\-\ln \left|5-B\right|=4t + D

If we want to find the explicit general solution of the differential equation

We isolate B

-\ln \left|5-B\right|=4t + D\\\ln \left|5-B\right|=-4t+D\\\left|5-B\right|=e^{-4t+D}

Recall the definition of |x|

|x|=\left \{ {{x, \:if \>x\geq \>0 } \atop {-x, \:if \>x0}} \right.

So

\left|5-B\right|=e^{-4t+D}\\5-B= \pm \:e^{-4t+D}\\B=5 \pm \:e^{-4t+D}\\B=5\pm \:e^{-4t}\cdot e^{D}\\B=5+Ae^{-4t}

where A=\pm e^{D}

Now B(1) =30 implies

B=5+Ae^{-4t}\\30=5+Ae^{-4}\\30-5=Ae^{-4}\\25e^{4}=A

And the solution is

B=5+(25e^{4})e^{-4t}\\B=5+25e^{-4t+4}

8 0
4 years ago
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