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3 years ago

PLEASEEE Find the angles

Mathematics

2 answers:

4vir4ik [10]3 years ago

4 0

Answer:

A= 58 degrees B=25 degrees

Step-by-step explanation:

A+B=83

(9x+4) + (4x+1)=83

x=6

Solve for angles

shtirl [24]3 years ago

3 0

Answer:

angel A = 25°

angel B = 58°

Step-by-step explanation:

ACD + ACB = 180°

83° + ACB = 180°

ACB = 180° - 83°

ACB = 97°

ACB + angel B + angel A = 180°

97° + 4x + 1° + 9x + 4° = 180°

13x = 180° - 102°

13x = 78°

x = 78°/13

x = 6

angel A = 4x + 1°

angel A = 4(6) + 1°

angel A = 24 + 1°

angel A = 25°

angel B = 9x + 4°

angel B = 9(6) + 4°

angel B = 54 + 4°

angel B = 58°

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(a) Take the Laplace transform of both sides:

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

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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