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just olya [345]
3 years ago
9

3 + 4 + 7 = 14 multiplication

Mathematics
2 answers:
SSSSS [86.1K]3 years ago
8 0
Multiplication:
7×2=14.
Fittoniya [83]3 years ago
4 0
If I understand correctly, you multiply 3•4•7=184
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Write the equations for the graphs below (worth 5 points) ​
serg [7]
Answers:

10) y= 1/2x - 2
11) y= 2x + 3
12) y= 2/3x - 4

I found this by using y=mx+ b
3 0
2 years ago
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A box of donuts containing 6 maple bars, 3 chocolate donuts, and 3 custard filled donuts is sitting on a counter in a work offic
Svetlanka [38]

Probabilities are used to determine the chances of selecting a kind of donut from the box.

The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068

The given parameters are:

\mathbf{Bars = 6}

\mathbf{Chocolate = 3}

\mathbf{Custard= 3}

The total number of donuts in the box is:

\mathbf{Total= 6 + 3 + 3}

\mathbf{Total= 12}

The probability of eating a chocolate donut, and then a custard filled donut is calculated using:

\mathbf{Pr = \frac{Chocolate}{Total}\times \frac{Custard}{Total-1}}

So, we have:

\mathbf{Pr = \frac{3}{12}\times \frac{3}{12-1}}

Simplify

\mathbf{Pr = \frac{3}{12}\times \frac{3}{11}}

Multiply

\mathbf{Pr = \frac{9}{132}}

Divide

\mathbf{Pr = 0.068}

Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068

Read more about probabilities at:

brainly.com/question/9000575

7 0
2 years ago
Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.
jarptica [38.1K]
  • y''-y'+y=\sin x

The corresponding homogeneous ODE has characteristic equation r^2-r+1=0 with roots at r=\dfrac{1\pm\sqrt3}2, thus admitting the characteristic solution

y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x

For the particular solution, assume one of the form

y_p=a\sin x+b\cos x

{y_p}'=a\cos x-b\sin x

{y_p}''=-a\sin x-b\cos x

Substituting into the ODE gives

(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x

-b\cos x+a\sin x=\sin x

\implies a=1,b=0

Then the general solution to this ODE is

\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}

  • y''-3y'+2y=e^x\sin x

\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2

\implies y_c=C_1e^x+C_2e^{2x}

Assume a solution of the form

y_p=e^x(a\sin x+b\cos x)

{y_p}'=e^x((a+b)\cos x+(a-b)\sin x)

{y_p}''=2e^x(a\cos x-b\sin x)

Substituting into the ODE gives

2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x

-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x

\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12

so the solution is

\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}

  • y''+y=x\cos(2x)

r^2+1=0\implies r=\pm i

\implies y_c=C_1\cos x+C_2\sin x

Assume a solution of the form

y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)

{y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)

Substituting into the ODE gives

(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)

-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)

\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49

so the solution is

\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}

7 0
3 years ago
Solve using elimination.<br> 3x − 5y = 18 <br> -10x + 5y = 10
olga55 [171]

3x - 5y = 18 over

-10x + 5y = 10

Add them because the 5y's have the right symbols for us to add

-7x = 28

Divide

x = -4

Now you can plug in -4 for x in one equation, I would use the first equation!

3(-4) - 5y = 18

-12 - 5y = 18

Add 12

-5y = 30

Divide

y = -6

Your solutions are going to be:

x = -4

y = -6

To check your work plug x and y into one equation:

3(-4) - 5(-6) = 18

-12 + 30 = 18

18 = 18

Since 18 does equal 18 you know that your solution's work!

5 0
2 years ago
Read 2 more answers
Grade low in math help if you can :) Geometry
mestny [16]

Answer:

C) 10 inches

Step-by-step explanation:

I think the simplest way to do this is as follows:

divide 75 by 15 to get how many times you need to multiply 2in by

75/15=5

since 15(5)=75 or the actual size of the airplane,

2(5)=10 which is the size of the model airplane

5 0
2 years ago
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