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hoa [83]
3 years ago
12

FAST 20 POINTS PLEASE HELP WILL GIVE BRAINIEST ANSWER What is the solution of the system of equations? {3x−1/2y=−7/2 x+2y=1 Ente

r your answer in the boxes. ( , )
Mathematics
1 answer:
avanturin [10]3 years ago
7 0
The answer is 17/10 and 82/10
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Trigonometry Problem, only solve if you know how and can tell me how to solve it.
Crazy boy [7]

The distances from the point the plane leaves the ground are given by the

trigonometric relationships of right triangle and Pythagoras theorem.

  • A) The minimum distance to the base of the tower is approximately <u>874.57 ft</u>.
  • B) The minimum distance to the top of the tower is approximately <u>882.76 ft</u>.

Reasons:

A) The angle with which the airplane climbs, θ = 11°

Height of the tower which the airplane flies, T = 120 foot

The clearance between the tower and the airplane, C = 50 feet

Required:

The minimum distance between the point where the plane leaves the ground and the base of the tower, \displaystyle d_{min}}

Solution:

Height at which the plane flies over the tower, h = T + C

Therefore, h = 120 ft. + 50 ft. = 170 ft.

At the point the plane leaves the ground, we have;

  • \displaystyle tan(\theta) = \mathbf{\frac{h}{d_{min}}}

Which gives;

\displaystyle tan(11^{\circ}) = \frac{170 \, ft.}{d_{min}}

\displaystyle d_{min} = \mathbf{\frac{170 \, ft.}{tan(11^{\circ})}} \approx 874.57 ft.

  • The minimum distance between the point where the plane leaves the ground and the base of the tower, \displaystyle d_{min}} ≈ <u>874.57 ft</u>.

B) The minimum distance between the point where the plane leaves the ground and the tower, <em>R</em>, is given by Pythagoras's theorem as follows;

R² = \displaystyle \mathbf{d_{min}}}² + T²

Which gives;

R = √(*874.57 ft.)² + (120 ft.²)) ≈ 882.76 ft.

  • The distance from the point where the airplane leaves the ground to the tower, R ≈ <u>882.76 ft</u>.

Learn more about Pythagoras theorem here:

brainly.com/question/11256912

7 0
2 years ago
Enter the fractions in order from the least to greatest 1/5 8/10 1/2
UkoKoshka [18]

Answer:

1/5, 1/2, 8/10

Step-by-step explanation:

If you change all the fractions into x/10, you get:

1/5 = 2/10

1/2 = 5/10

8/10 = 8/10

2/10 < 5/10 < 8/10

hope this helps :)

6 0
4 years ago
Last week 3,000 people came to hear a presidential candidate speak in Central City. If they represent 162/3 of the people who li
RSB [31]

3000 = 162/3

162/3 = 54

3000 = 54%

3000 * 0.54 =

1620 Hope I've helped!

7 0
3 years ago
Position vectors m and n have terminal points of (2, ) and (1, y), respectively.
OLEGan [10]
This problem involves the dot product.

You must provide all info for position vector m.  Your (2,) is inadequate.

Supposing that the terminal point of vector m were (2,3), then

m dot n would equal 8, or 8 = 2*2+3*y.  Then 8 = 4 + 3y, and 4 = 3y, and y =3/4.

Please type in the terminal point of vector m and then answer this question following the above example.

7 0
3 years ago
Find an equation in standard form for the hyperbola with vertices at (0,plus or minus 6) and asymptotes at y= (plus or minus 3/5
antoniya [11.8K]

Answer:

\frac{(y)^2}{36}-\frac{(x)^2}{100}=1

Step-by-step explanation:

Given:

Vertices of Hyperbola : (0 ± 6) or (0,6) and (0,-6)

and asymptotes at y= (±3/5)x 0r y= 3/5 x and y=-3/5 x

The vertices are of vertical hyperbola. The equation used will be:

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\\

The Center of hyperbola (h,k) =(0,0)

The Distance from vertices to center is a and a = 6 (given)

For equation we have value of h,k and a and need to find value of b

we know,

y= k ± a/b (x-h)

Values of h and k are zero

y= 0 ± a/b (x-0)

y= (± a/b ) x

We are given asymtotes at y= (± 3/5)x which is equal to y= (± a/b ) x

as a = 6 then b= 10 i.e The simplified form of 6/10 is 3/5 so value of b=10

Putting values of a,b,h and k in equation we get,

\frac{(y-0)^2}{(6)^2}-\frac{(x-0)^2}{(10)^2}=1\\\\\frac{(y)^2}{36}-\frac{(x)^2}{100}=1

7 0
4 years ago
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