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strojnjashka [21]
3 years ago
12

If anyone is good at right triangle trig word problems please help

Mathematics
2 answers:
yaroslaw [1]3 years ago
6 0

Answer:

  • 91 feet

Step-by-step explanation:

<em>Refer to attached picture</em>

Let the base is b and the height of the hill is x.

<u>For the smaller triangle we have:</u>

  • x/b = tan 39.3°

<u>And for the bigger triangle: </u>

  • (x + 40)/b = tan 49.7°

<u>Solve both equations for b:</u>

  • b = x/ tan 39.3°
  • b = (x + 40)/tan 49.7°

<u>Eliminate b and get the values of tan, solve for x:</u>

  • x/ tan 39.3° = (x + 40) / tan 49.7°
  • x tan 49.7° = (x + 40) tan 39.3°
  • x( 1.179) = x(0.819) + 40(0.819)
  • x(1.179 - 0.819) = 32.76
  • x = 32.76/0.36
  • x = 91 feet (rounded to the nearest foot)

erastovalidia [21]3 years ago
3 0

Answer:

The height of the hill is 131ft

See explanation in the photo

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What number should be added to 7/12 to get 4/15?​
DENIUS [597]

Answer:

Step-by-step explanation:

Let the number = x.

We are trying to find out what number (x) we would need to add to 7/12 to get 4/15

\frac{7}{12}  + x = \frac{4}{15}

Subtract 7/12 from both sides.

x = \frac{4}{15}  - \frac{7}{12}

We need to make the denominators of the fractions the same.

\frac{4}{15}  = \frac{16}{60}

\frac{7}{12}  = \frac{35}{60}

So,

x = \frac{16}{60}  - \frac{35}{60}  = -  \frac{19}{60}

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A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the
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Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring \frac{8}{3} feet.

Mass=m=\frac{W}{g}

Mass=m=\frac{16}{32}

g=32 ft/s^2

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By hook's law

w=kx

16=\frac{8}{3} k

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f(t)=10cos(3t)

A damping force is numerically equal to 1/2 the instantaneous velocity

\beta=\frac{1}{2}

Equation of motion :

m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)

Using this equation

\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)

\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)

\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)

Auxillary equation

m^2+m+12=0

m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}

m=\frac{-1\pmi\sqrt{47}}{2}

m_1=\frac{-1+i\sqrt{47}}{2}

m_2=\frac{-1-i\sqrt{47}}{2}

Complementary function

e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})

To find the particular solution using undetermined coefficient method

x_p(t)=Acos(3t)+Bsin(3t)

x'_p(t)=-3Asin(3t)+3Bcos(3t)

x''_p(t)=-9Acos(3t)-9sin(3t)

This solution satisfied the equation therefore, substitute the values in the differential equation

-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)

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Comparing on both sides

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3B-3A=0

Adding both equation then, we get

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Substitute the value of B in any equation

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Particular solution, x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

Now, the general solution

x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)

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x(0)=2 ft

x'(0)=0

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2-\frac{10}{3}=c_1

c_1=\frac{-4}{3}

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c_2==-\frac{64}{3\sqrt{47}}

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