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3 years ago

12

Help plz I'm desperate

1 answer:

lutik1710 [3]3 years ago

7 0

<u>Anna:</u>

1st month = 18 + $\frac{3}{4}$ = 18 $\frac{3}{4}$ inches

2nd month = 18 $\frac{3}{4}$ + $\frac{3}{4}$ = 19 $\frac{1}{2}$

3rd month = 19 $\frac{1}{2}$ + $\frac{3}{4}$ = 20 $\frac{1}{4}$

4th month = 20 $\frac{1}{4}$ + $\frac{3}{4}$ = 21

5th month = 21 + $\frac{3}{4}$ = 21 $\frac{3}{4}$

<u>Grace:</u>

Because Anna is $\frac{9}{4}$ inches shorter than Grace, we can start the count from the third month:

1st month = 20 $\frac{1}{4}$

2nd month = 21

3rd month = 21 $\frac{3}{4}$

4th month = 21 $\frac{3}{4}$ + $\frac{3}{4}$ = 22 $\frac{1}{2}$

5th month = 22 $\frac{1}{2}$ + $\frac{3}{4}$ = 23

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3 years ago

During an algebra class a teacher asked for examples of non-prime numbers. Jake says 51, Teresa says 41, and let Liu says 21. Wh

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3 years ago

Read 2 more answers

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

$\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}$

$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

$\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13$

So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

Integrating both sides yields

$\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx$

$\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C$

$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

With the initial condition y(0) = 1, we find

$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

so that the particular solution is

$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

It's not too hard to solve explicitly for y; notice that

$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

$\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}$

$y+1 = \dfrac6{2+e^{3x}}$

$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

7 0

2 years ago