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2 years ago

Which is the graph of f(x) = 2(3)^x

Mathematics

1 answer:

creativ13 [48]2 years ago

4 0

Answer:

Step-by-step explanation: The graph is vertically stretched by a scale factor of 0.5 and shifted up 3 units.

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Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

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3 years ago

Which set of angles could be the interior angles of a triangles

qwelly [4]

Answer:

Step-by-step explanation:

6 0

2 years ago

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Dvinal [7]

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

What is the equation of the graphed line written in

Ghella [55]

Answer:

X-Y=-3

Step-by-step explanation:

6 0

3 years ago

Let f(x) = 4(x^2) - 3x

posledela

Answer:

(f+g)(x)=5x²-4x+3

(f-g)(x)=3x²-2x+3

(fg)(x) $=4x^4-7x^3+15x^2-9x$

$\frac{f}{g}(x)$ $=\frac{4x^2-3x}{x^2-x+3}$

Step-by-step explanation:

Given that,

f(x)=4x²-3x

g(x)=x²-x+3

(f+g)(x)

=f(x)+g(x)

=4x²-3x+x²-x+3

=(4x²+x²)+(-3x-x)+3 [ combined the like terms]

=5x²-4x+3

(f-g)(x)

=f(x)-g(x)

=4x²-3x-(x²-x+3)

=4x²-3x-x²+x-3

=(4x²-x²)+(-3x+x)-3 [ combined the like terms]

=3x²-2x+3

(fg)(x)

=f(x).g(x)

=(4x²-3x).(x²-x+3)

=4x²(x²-x+3)-3x(x²-x+3)

$=4x^4-4x^3+12x^2-3x^3+3x^2-9x$

$=4x^4+(-4x^3-3x^3)+(12x^2+3x^2)-9x$

$=4x^4-7x^3+15x^2-9x$

$\frac{f}{g}(x)$

$=\frac{f(x)}{g(x)}$

$=\frac{4x^2-3x}{x^2-x+3}$

7 0

3 years ago

Which is the graph of f(x) = 2(3)^x​

Which is the graph of f(x) = 2(3)^x