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jekas [21]
3 years ago
14

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
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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
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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
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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
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