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Rashid [163]
3 years ago
5

The graphs of f(x) and g(x) are shown below.

Mathematics
2 answers:
nexus9112 [7]3 years ago
7 0

Answer:

On a coordinate plane, a straight line with a positive slope goes through points (negative 2, negative 6), (0, 0), and (2, 6).

Step-by-step explanation:

The function is given by f(x) = - x and another function is given by g(x) = 2x.

Now, (g - f)(x) will be given by

(g - f)(x) = 2x - (- x) = 3x .......... (1)

Now, we have to choose from the options of the graphs that is applicable to function (1).

So, on a coordinate plane, a straight line with a positive slope goes through points (negative 2, negative 6), (0, 0), and (2, 6). (Answer)

Alex73 [517]3 years ago
7 0

Answer:

Option C

Step-by-step explanation:

We are given that

f(x)=-x

The line passing through the points (0,0),(-6,6) and (6,-6).

g(x)=2x

The line passing through the points (-3,-6),(0,0) and (3,6).

We have to find the graph of (g-f)(x).

(g-f)(x)=g(x)-f(x)=2x-(-x)=2x+x

(g-f)(x)=3x

Substitute x=0

(g-f)(0)=3(0)=0

(g-f)(2)=3(2)=6

(g-f)(-2)=3(-2)=-6

(g-f)(-6)=3(-6)=-18

Option C is true.

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Answer:

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y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

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y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

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