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vichka [17]
3 years ago
10

What is the slope of the line?

Mathematics
2 answers:
snow_tiger [21]3 years ago
7 0

Answer:

Y=3x-2

Step-by-step explanation:


damaskus [11]3 years ago
5 0

Answer: Slope = 3

             Function: y = 3x+6


Step-by-step explanation:


Gather 2 points.

(-1, 3), (-3, -3)

Divide: (y1 - y2) / (x1 - x2)

-3 -3 =-6

-3 -(-1) = -2

-6/-2 = 3


Slope is 3 (m = 3)

Point-Slop Intercept:


(y - y1) = m(x - x1)

y-(-3) = 3(x-(-3))

       (two negatives = Postive) Distribute the slope to the numbers inside parenthesis:

y-(-3) = 3x + 9

y + 3 = 3x + 9

Subtract 3 from both sides

y = 3x + 6.

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A group of 485 people take a canoe trip. They fill up all avalible canoes that each hold 3 people with 273 people. The rest of t
Jobisdone [24]
Answer:
3c=273
c=91
2c= 212
C= 106
c= 91+ 106= 197
so they will need 197 canoes total

Why:
So first you have to determine a set variable to represent the number of canoes, I chose C. Then you make an equation to represent the number of canoes 273 people will use if they group into 3's, from this I got 3c=273. Solve for C and get 91.
The remainder of the group which is 485-273= 212 will use canoes in groups of 2's. To represent this, 2c=212. Solve for C and get 106. Combine 106 and 91 to get the total number of canoes.
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Answer:

False

Step-by:-step explanation

2 million years

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Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.
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  • 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}

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\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)}

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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)

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so the solution is

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

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