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Damm [24]
3 years ago
13

In ΔFGH, f = 35 inches, g = 20 inches and h=30 inches. Find the measure of ∠H to the nearest degree.

Mathematics
2 answers:
kogti [31]3 years ago
5 0

Answer:

100

Step-by-step explanation:

it just is

svp [43]3 years ago
3 0

9514 1404 393

Answer:

  59°

Step-by-step explanation:

The law of cosines tells you ...

  h² = f² +g² -2fg·cos(H)

Solving for angle H gives ...

  H = arccos((f² +g² -h²)/(2fg)) = arccos(725/1400) ≈ 58.811°

  ∠H ≈ 59°

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3 years ago
Consider the initial value problem y′+2y=4t,y(0)=8.
Xelga [282]

Answer:

Please read the complete procedure below:

Step-by-step explanation:

You have the following initial value problem:

y'+2y=4t\\\\y(0)=8

a) The algebraic equation obtain by using the Laplace transform is:

L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\

next, you replace (1) and (2):

sY(s)-y(0)+2Y(s)=\frac{4}{s^2}\\\\sY(s)+2Y(s)-8=\frac{4}{s^2}  (this is the algebraic equation)

b)

sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)} (this is the solution for Y(s))

c)

y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}

To find the inverse Laplace transform of the first term you use partial fractions:

\frac{4}{s^2(s+2)}=\frac{-s+2}{s^2}+\frac{1}{s+2}\\\\=(\frac{-1}{s}+\frac{2}{s^2})+\frac{1}{s+2}

Thus, you have:

y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}  

(this is the solution to the differential equation)

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Step-by-step explanation:

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