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aniked [119]
3 years ago
15

Alice and her friend Emma leave on different flights from the same airport. Alice's flight flies 100 miles due south, then turns

70° toward west and flies 50 miles. Emma's flight flies 100 miles due north, then turns 50° toward east and flies 50 miles. Which fight among you is farther from the airport? Explan your rreasoning
don't spam!
Mathematics
1 answer:
DaniilM [7]3 years ago
3 0

9514 1404 393

Answer:

  Emma's flight is farther from the airport

Step-by-step explanation:

Even without doing any detailed calculation, you can see on a graph that the larger turn Alice's flight makes keeps it closer to the airport.

Emma's flight is farther from the airport.

__

The distance can be calculated using the Law of Cosines. The angle of interest is the supplement of the turn angle. Since the leg distances are the same in each case, the only variable affecting the distance from the airport is the angle measure. Again, the larger the turn, the shorter the distance from the airport. If T is the turn angle, the distance is ...

  d² = 100² +50² -2·100·50·cos(180° -T)

  d² = 12500 +10000·cos(T) . . . . . . . using cos(180°-T) = -cos(T)

  d = 100√(1.25+cos(T))

For Alice's 70° turn, the distance is ...

  d = 100√(1.25+0.342) ≈ 126.18 mi

For Emma's 50° turn, the distance is ...

  d = 100√(1.25 +0.643) ≈ 137.58 mi

Emma's flight is farther from the airport.

_____

The law of cosines formula is ...

  c² = a² +b² -2ab·cos(C)

where triangle sides are lengths a, b, c and angle C is opposite side c.

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A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta
nalin [4]

Answer:

Option 2 - Approximately 24–36 pounds

Step-by-step explanation:

Given : A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of standard Eskimo dogs are normally distributed.

To find : What range of weights would 99.7% of the dogs have?

Solution :

The range of 99.7% will lie between the mean ± 3 standard deviations.

We have given,

Mean weight of Eskimo dogs is \mu=30

Standard deviation of Eskimo dogs is \sigma=2

The range of weights would 99.7% of the dogs have,

R=\mu\pm3\sigma

R=30\pm3(2)

R=30\pm6

R=30+6,30-6

R=36,24

Therefore, The range is approximately, 24 - 36 pounds.

So, Option 2 is correct.

5 0
4 years ago
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3 years ago
The maths question for algebra is
ankoles [38]

Step-by-step explanation:

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4 0
3 years ago
Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7
Digiron [165]
y=x^r
\implies r(r-1)x^r+6rx^r+4x^r=0
\implies r^2+5r+4=(r+1)(r+4)=0
\implies r=-1,r=-4

so the characteristic solution is

y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}

As a guess for the particular solution, let's back up a bit. The reason the choice of y=x^r works for the characteristic solution is that, in the background, we're employing the substitution t=\ln x, so that y(x) is getting replaced with a new function z(t). Differentiating yields

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}
\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)

Now the ODE in terms of t is linear with constant coefficients, since the coefficients x^2 and x will cancel, resulting in the ODE

\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t

Of coursesin, the characteristic equation will be r^2+6r+4=0, which leads to solutions C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}, as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If z_1,z_2 are the solutions to the characteristic equation of the ODE in terms of z, then we can find another of the form z_p=u_1z_1+u_2z_2 where

u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt
u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt

where W(z_1,z_2) is the Wronskian of the two characteristic solutions. We have

u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t

u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t

\implies z_p=u_1z_1+u_2z_2
\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t

and recalling that t=\ln x\iff e^t=x, we have

\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x
4 0
3 years ago
Grayson is flying a kite, holding his hands a distance of 2.75 feet above the ground and letting all the kite’s string play out.
sergiy2304 [10]

The position of the kite with the point directly beneath the kite at the same

level with the hand and the hand for a right triangle.

  • The height of the kite above the is approximately <u>66.314 feet</u>.

Reasons:

The height of his hands above the ground, h = 2.75 feet

Angle of elevation of the string (above the horizontal), θ = 26°

Length of the string, <em>l</em> = 145 feet

Required:

The height of the kite above the ground.

Solution:

The height of the kite above the ground is given by trigonometric ratios as follows;

\displaystyle Height \ of \ kite, \, k_h = \mathbf{h + l \times sin(\theta)}

Therefore;

\displaystyle Height \ of \ kite, \, k_h = 2.75 + 145 \times sin(26^{\circ}) \approx \mathbf{66.314}

The height of the kite above the, k_h ≈ <u>66.314 feet</u>

Learn more about trigonometric ratios here:

brainly.com/question/9085166

8 0
3 years ago
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