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nikklg [1K]
3 years ago
5

Find the value of x in the figure below x = degrees x = degrees

Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
6 0

Answer:

x = 190 and x = 126

Step-by-step explanation:

In both diagrams the indicated angles are vertical and congruent, then

\frac{x}{2} = 95 ( multiply both sides by 2 )

x = 190

and

x + 21 = 147 ( subtract 21 from both sides )

x = 126

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Agata [3.3K]

Answer:

Part A

5yr old boy.

Y=-(1/8)x+113/8 x=5

Y=-5/8+113/8=(113-5)/8=108/8

Y=13.5 hrs

Yes the estimated value of sleep by this function is reasonable as it is close to childhood more sleep is likely.

Part B

20yr old boy.

Y=-(1/8)x+113/8 x=20

Y=-20/8+113/8=(113-20)/8=93/8

Y=11.63 hrs

No the estimated value of sleep by this function is not reasonable as the normal sleep of 7-8 hrs is enough for adults.

Step-by-step explanation:

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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

Mass,m=\frac{1}{2} Slug

By hook's law

w=kx

16=\frac{8}{3} k

k=\frac{16\times 3}{8}=6 lb/ft

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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Adding both equation then, we get

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Now, the general solution

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x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

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

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

c_1=\frac{-4}{3}

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\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0

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