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2 years ago

12

A car dealership sold the following colors of cars in January.

1 answer:

sweet [91]2 years ago

7 0

The answer is A. 28%

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Please help me with this. This problem is actually making my life worst at the moment. Please, someone, help me.

Xelga [282]

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3 years ago

PLZ HELP!! I NEED QUICKLY!!

Amiraneli [1.4K]

Answer:

(1, 1), (2, 2), (-3, -3), (4, 4), (-5, -5)

Step-by-step explanation:

You get a straight line. As you can see in the picture, the figure lies in the first and third quadrant.

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2 years ago

Read 2 more answers

3.249 rounded to nearest tenth

Viktor [21]

Answer:

3.2

Step-by-step explanation:

4 is in the hundredths place. It is less than 5 so you leave the tenths place alone, so you get 3.2.

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3 years ago

Read 2 more answers

10x + 7y=1<br> -5x-7y=24

nikitadnepr [17]

Okay, so since both y values share the same number, let's add both equations so the y values cancel out leaving us to solve for x.

10x+(-5x)=5x

7y+(-7y)=0

1+24=25

Now we have 5x=25

divide both sides by 5 and we have x=5

Now that we know x=5, we can plug it into one of the problems to find y.

Let's do 10(5) +7y=1

50+7y=1 subtract 5 and we get 7y=-49

divide -7 from both sides and we get y=-7

so there you have it x=5 and y=-7

8 0

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

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

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

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

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

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

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

7 0

2 years ago