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2 years ago

8

Your friend is 1.2 times taller than you. Your friend is 64.5 inches tall. How tall are you?

1 answer:

Andru [333]2 years ago

6 0

You are X, if your friend is 1.2 times taller than you= 1.2X
That's why 64.5=1.2X
X=53.75
But you should write 53.8 inches tall. Because there is one number after point.
<span>As a result, you are 53.8 inches tall

</span>

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Step-by-step explanation:

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Use the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ to calculate the length of each segment.

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✔️Distance between X(1, 2) and Y(2, -4)

Let,

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Plug in the values

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Let,

$Y(2, -4) = (x_1, y_1)$

$Z(-2, -1) = (x_2, y_2)$

Plug in the values

$YZ = \sqrt{(-2 - 2)^2 + (-1 -(-4))^2}$

$YZ = \sqrt{(-4)^2 + (3)^2}$

$YZ = \sqrt{16 + 9}$

$YZ = \sqrt{25}$

$YZ = 5$

✔️Distance between Z(-2, -1) and W(-1, 1)

Let,

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Plug in the values

$ZW = \sqrt{(-1 -(-2))^2 + (1 - (-1))^2}$

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3 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

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

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

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

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

So we have

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

Integrating both sides yields

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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