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Genrish500 [490]
3 years ago
5

A kite is stuck in a 36-ft tree. If the angle of elevation from the kite and the ground is 12, find the length of the kite.

Mathematics
1 answer:
yanalaym [24]3 years ago
4 0

Answer:

173.15 ft.  

Step-by-step explanation:

Please find the attachment.

Let L be the length of kite.  

We have been given that a kite is stuck in  36-ft tree. The angle of elevation from the kite and the ground is 12.

We can see from our attachment that kite and tree form a right triangle. height of tree is opposite of angle of elevation and length of kite is hypotenuse.

Since we know that Sine relates the opposite and hypotenuse of a right triangle, so we will use sine to find the length of kite.

sin(12)=\frac{36}{L}    

L=\frac{36}{sin(12)}

L=\frac{36}{0.207911690818}  

L=173.1504364105882792\approx 173.15

Therefore, the length of kite is 173.15 ft.

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Determine whether each of the following sequences are arithmetic, geometric or neither. If arithmetic, state the common differen
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\qquad\qquad\huge\underline{{\sf Answer}}

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\textsf{Since the common ratio is same, } \textsf{we can infer that it's a geometric progression} \textsf{with common ratio of -3}

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3 years ago
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
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Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

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(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

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\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

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Jessica purchased a DVD that was on sale for 12% off.

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Here, we will form an algebraic expression, according to the given data.

So,

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Learn more about sales tax here:brainly.com/question/9437038

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