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3 years ago

5

A classroom has 8 boys and 14 students in the whole class whats the ratio from boys to girls

2 answers:

Mariana [72]3 years ago

3 0

8/14 boys, 6/14 girls which means 4/7 boys and 3/7 girls

NeX [460]3 years ago

3 0

The ratio would be 8:6

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4 years ago

What is the slope of the line that passes through the points ( 8 , − 6 ) and (8,−9)? Write your answer in simplest form.

Natalka [10]

Answer:

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2 years ago

If a car with an average speed of 40 miles per hour drives for 45 minutes how far will it have gone

REY [17]

Answer: $30\ miles$

Step-by-step explanation:

The average speed of the car is 40 miles per hour:

$V=40\ \frac{mi}{h}$

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$t=45\ min$

Since 1 hours has 60 minutes, we can convert the time from minutes to hours:

$(45\ min)(\frac{1\ h}{60\ min})=0.75\ h$

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Therefore, substituting values into the formula, we get:

$d=(40\ \frac{mi}{h})(0.75\ h)\\\\d=30\ mi$

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4 years ago

After all of the start-up costs, a company starts with $100 and makes $0.75 on each unit sold. Write a linear equation in slope

KiRa [710]

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3 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

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

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

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

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

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

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

7 0

2 years ago