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Svetllana [295]
3 years ago
5

Will and olly share £80 in the ratio 3 : 2

Mathematics
1 answer:
Lelu [443]3 years ago
7 0

Answer:

Will's share =   48

Olly's share =  32

Step-by-step explanation:

Will's share : Olly's share = 3:2

Will's share = 3x

Olly's share = 2x

3x + 2x =  £ 80                 {Combine like terms}

     5x   = 80                    {Divide both sides by 5}

      x = 80/5

       x = 16

Will's share = 3x = 3*16 = 48

Olly's share = 2x = 2*16 = 32

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

Answer:

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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}

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