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sergiy2304 [10]
3 years ago
12

Manuel and Peter are standing 12 ft. apart from each other and looking up at a kite that is flying between them. Manuel is flyin

g the kite on a 65 ft. string at an angle of 70(degrees) with the ground. How far is Peter from the kite?
A. 61.9 ft.
B. 65.8 ft.
C. 68.1 ft.
D. 99.5 ft.
Mathematics
1 answer:
Viktor [21]3 years ago
4 0

Answer:

C. 68.1 ft.

Step-by-step explanation:

Hope this helps

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Find x so that (x,-14) is a solution to 2x+3y=12
Monica [59]

Answer:

27

Question:

Find x so that (x,-14) is a solution to 2x+3y=12 .

Step-by-step explanation:

So we are asked to replace y with -14 and solve for x in the equation:

2x+3y=12.

2x+3y=12 with y=-14:

2x+3(-14)=12

2x-42=12

Add 42 on both sides:

2x=54

Divide both sides by 2:

x=27.

So the x-coordinate that corresponding to y=-14 is 27.

Verify that:

(27,-14) is a point satisfying 2x+3y=12.

Replace x with 27 and y with -14:

2(27)+3(-14)=12

54+-42=12

12=12 is true so our work must be right.

3 0
3 years ago
Find the area of the rhombus.<br> 7.5 yd<br> 13 yd<br> 13 yd<br> 7.5 yd<br> [ ? ] yd2
Doss [256]

Answer:

195

Step-by-step explanation:

4 0
2 years ago
The sunrise café gets tea bags of 1,000. If the café charges $1.75 for each cup of tea, and each cup of tea gets 1 tea bag, how
Setler [38]

Answer:

You would do $1.75 x 1,000. That would be 1$,750. Then, you would take away that $95 because they spent buying the box of tea bags, and their total would be $1,750 - $95 = $1,655.

5 0
2 years ago
A basin is filled by two pipes in 12 minutes and 16 minutes respectively. Due to the obstruction of water flow after the two pip
olasank [31]

Answer:

The time duration of the two pipes restricted flow before the flow became normal is 4.5 minutes

Step-by-step explanation:

The given information are;

The time duration for the volume, V, of the basin to be filled by one of the pipe, A, = 12 minutes

The time duration for the volume, V, of the basin to be filled by the other pipe, B, = 16 minutes

Therefore, the flow rate of pipe A = V/12

The flow rate of pipe B = V/16

Due to the restriction, we have;

The proportion of its carrying capacity the first pipe, A, carries = 7/8 of the carrying capacity

The proportion of its carrying capacity the second pipe, B, carries = 5/6 of the carrying capacity

Whereby the tank is filled 3 minutes after the restriction is removed, we have;

\dfrac{7}{8} \times \dfrac{V}{12} \times t + \dfrac{5}{6} \times \dfrac{V}{16} \times t +  \dfrac{V}{12} \times 3 + \dfrac{V}{16} \times 3 = V

Simplifying gives;

\dfrac{(2\cdot t +7) \cdot V}{16}  = V

2·t + 7 = 16

t = (16 - 7)/2 = 4.5 minutes

Therefore, it took 4.5 seconds of the restricted flow before the the flow of water in the two pipes became normal

7 0
3 years ago
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Solve the system of inequalities bu graphing and shading: y&lt;2x-3 and 2x+y&gt;2
Mars2501 [29]

Answer:

Step-by-step explanation:

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3 years ago
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