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Masteriza [31]
3 years ago
9

A 12-foot flagpole casts a nine-foot shadow. What is the measure from the top of the shadow to the top of the flag pole?

Mathematics
2 answers:
ziro4ka [17]3 years ago
8 0

Answer:

15

Step-by-step explanation:

I used common sense.

amm18123 years ago
5 0

Here is the answer OwO:

A 12-foot flagpole casts a nine-foot shadow. The measure from the top of the shadow to the top of the flag pole is 15.

Here is the explanation/work OwO:

<u><em>Switch sides.</em></u>

c^2=12^2+9^2

<u><em>12^2=144</em></u>

c^2=144+9^2

<u><em>9^2=81</em></u>

c^2=144+81

<u><em>Add the numbers: 144 + 81 = 225</em></u>

c^2=225

<u><em>For x^2 = f(a) the solution is √f(a)</em></u>

c = 15

Therefore, your answer will be 15.

_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-

Sorry if I am late. Hope this helps!! ^^

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Step-by-step explanation:

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The mean is the average of a set of numbers.  So to do that:


add all the numbers 120+235+98+142=$595

Then divide that by the total number in the set, in this case it is 4.

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What is the least common multiple (LCM) of 6 and 10?<br> OA. 2.<br> OB. 20<br> O C. 30<br> OD. 60
Naily [24]

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Step-by-step explanation:

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A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739
OlgaM077 [116]

Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

  1. For a constant acceleration: v_{f}=v_{0}+at, where  v_{f} is the final velocity in a direction after the acceleration is applied, v_{0} is the initial velocity in that direction  before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
  2. <em>Then for the x direction</em> it is known that the initial velocity is v_{0x} = 5320 m/s, the acceleration (the applied by the engine) in x direction is a_{x} 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the  is 739 s. Then: v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}
  3. In the same fashion, <em>for the y direction</em>, the initial velocity is  v_{0y} = 0 m/s, the acceleration in y direction is a_{y} 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}
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