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3 years ago

Plz answer these questions without putting links and files

Mathematics

2 answers:

kotykmax [81]3 years ago

8 0

Answer:

11/8 and 21/16

Step-by-step explanation:

3 x 1 = 3

4 x 2= 8

3/8+3/4(convert to eighth's)

3/8+6/8=11/8

1 1/8 = 9/8

9 x 1 = 9

8 x 2 = 16

9/16+3/4(convert to sixteenth's)

9/16+12/16

21/16

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hope it helps

hoa [83]3 years ago

8 0

=3/4 × 1/2 + 3/4

=3/8 + 3/4

=9/8

2nd questions ans

=1×1/2 + 3/4

=1/2 + 3/4

=5/4

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4 years ago

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Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

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

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

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 $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

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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 $Y(s)$ :

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

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

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

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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