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ollegr [7]
3 years ago
5

Assume a plane is flying directly north at 200 mph, but there is a wind blowing west at 23 mph. Part I: Express both the velocit

y of the plane and the velocity of the wind as vectors, using proper notation to represent each direction of motion. Part II: What is the velocity vector of the plane? Part III: What is the ground speed of the plane?
Mathematics
1 answer:
yuradex [85]3 years ago
7 0
Define i  as a unit vector in the eastern direction.
Define j as a unit vector in the northern direction.

Part I
Because the wind is blowing west, its velocity vector is
 -23i mph or as (-23, 0) mph
Because the plane is traveling north, its velocity vector is
  200j mph or as (0, 200) mph 

Part II
The actual velocity of the plane is the vector sum of the plane and wind velocities.
That is,
200j - 23i or (-23, 200) mph

Part III
The ground speed of the plane is the magnitude of its vector.
The ground speed is
√[200² + (-23)²] = 201.32 mph

The ground speed of the plane is 201.3 mph (nearest tenth)

Not:
The direction of the plane is
 tan⁻¹ 23/200 = 6.56° west of north.


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Mr Smith's art class took a bus trip to an art museum. The bus averaged 65 miles per hour on the highway and 25 miles per hour i
Leya [2.2K]
Let x be the distance traveled on the highway and y the distance traveled in the city, so:
\left \{ {{x+y=375} \atop { \frac{1}{65}x+ \frac{1}{25}y =7}} \right.
 
Now, the system of equations in matrix form will be:
\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right]   \left[\begin{array}{ccc}x&\\y&\end{array}\right] =  \left[\begin{array}{ccc}375&\\7&\end{array}\right]

Next, we are going to find the determinant:
D=  \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}
Next, we are going to find the determinant of x:
D_{x} =  \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8

Now, we can find x:
x=  \frac{ D_{x} }{D} = \frac{8}{ \frac{8}{325} } =325mi

Now that we know the value of x, we can find y:
y=375-325=50mi

Remember that time equals distance over velocity; therefore, the time on the highway will be:
t_{h} = \frac{325}{65} =5hours
An the time on the city will be:
t_{c} = \frac{50}{25} =2hours

We can conclude that the bus was five hours on the highway and two hours in the city. 

8 0
3 years ago
Multiple Choice: Choose the correct simplified expression for (3x - y) (W+p-3).
murzikaleks [220]

Answer:

answer this 5-y this the answer thanks for

8 0
2 years ago
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John has 9 reports to grade. There are 6 pages for each report. He grades 12 pages. How many more does he have to grade?
Goshia [24]
6 x 9 = 54
He has a total of 54 pages to grade.
54 - 12 = 42 pages

He still has to grade 42 pages.

Hope I helped!!
5 0
3 years ago
(Right angle Trigonometry) help me solve for X please!
Vinvika [58]

Answer:

x = 52.0

Step-by-step explanation:

Reference angle = 68°

Length of side Opposite to reference angle = x

Adjacent length = 21

Apply the trigonometric function, TOA, which is given as:

Tan 68° = Opp/Adj

Substitute

Tan 68° = x/21

21 × Tan 68° = x

51.9768238 = x

x = 51.9768238 ≈ 52.0 (approximated to nearest tenth)

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Pls help. I don't understand.
Nana76 [90]

Answer:

112

Step-by-step explanation:

I took the same test

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