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3 years ago

7

Over four evenings, Ms.Cross grades 6 papers, 8 papers, 9 papers, and 7 papers. She wants to grade an average of 9 papers a nigh

t. What would be a reasonable estimate of how many papers Ms.Cross must grade on the fifth day to reach that goal?
A)11
B)15
C)18
D)19

1 answer:

cupoosta [38]3 years ago

7 0

The answer would be 15. The average of 6, 7, 8, 9, and 15 is 9. Hope this helps.

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3 years ago

1. tan a = cot(a + 10°)

Temka [501]

Answer:

$a=40+90n$

Step-by-step explanation:

Use a cofunction identity on the right hand side or left hand side...

So $\tan(a)=\cot(90-a)$ .

We have the equation:

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Make the above replacement:

$\cot(90-a)=\cot(a+10)$

Since cotangent has period 180 degrees, we can also write this as:

$\cot(90-a+180n)=\cot(a+10)$

So solving the following will give us a set of solutions for $a$ :

$90-a+180n=a+10$

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2 years ago

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

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

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

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

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

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

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2 years ago