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3 years ago

A dog walking service charges a $19 one-time fee, plus $40 per week. The

Mathematics

1 answer:

rusak2 [61]3 years ago

7 0

Answer:

Kevin's estimate is incorrect. When x=4, the value of y is $116, which is not close to

$180.

Step-by-step explanation:

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Plz help becuase im trying to have this done in one min . ☝☝

In-s [12.5K]

Answer:

yes

Step-by-step explanation:

they both multiply by 20

4 0

3 years ago

Read 2 more answers

The image on a movie poster was shrunk to make the DVD cover art for the movie, so that the cover art is a scale image of the po

Bond [772]

Please see the attached image for a visual representation of our scale factor. We can set up this proportion by taking the DVD cover and poster values and placing them in fractions. Cross multiply and divide to solve for x.

6 0

3 years ago

Read 2 more answers

In a triangle the measure of the first angle is twice the measure of the second angle. the measure of the third angle is 8 degre

Rufina [12.5K]

Let's call the missing angles Q, R, and S. We know that Q = 2R and S = R - 8 and we also know that Q + R + S = 180. So that really just means that 180 = 4R - 8 Add 8 on both sides of the equal sign
188 = 4R So we divide on both sides of the equal sign by 4
47 = R And that means that Q = 2 * 47 So Q = 94 and S = 47 - 8 So S = 39
39 + 94 + 47 = 180
The missing angles are 39 degrees 94 degrees and 47 degrees.

4 0

3 years ago

Hernandez has 17 tomato plants that she wants to plant in rows.she will put 2 plants in some rows and 1 plant in the others.how

RideAnS [48]

Seven, I think. That's what I found somewhere else.

3 0

3 years ago

Read 2 more answers

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago