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nata0808 [166]
2 years ago
14

Brianna needs to contact members of the softball league. She calls 4 members in the morning. Those 4 people each call 4 more peo

ple in the afternoon. That evening, those additional people each call 4 others. How many people are called that evening? ​
Mathematics
2 answers:
allsm [11]2 years ago
8 0

Answer:

64 people

Step-by-step explanation:

in the morning, 4 people are called. Those 4 people call 4 more people, which brings it to 16 people called. Those 16 people then each call 4 others, which is 16 x 4. Giving you the answer of 64 people for the evening.

FYI, if the question asked how many people were called in total, you would add the 4 people from the morning, 16 people from the afternoon, and 64 people in the evening to get a total of 84 people called.

Hope this helps!

charle [14.2K]2 years ago
4 0

Answer:

...

Step-by-step explanation:

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natima [27]

Answer:

Speed of the plane in still air: 779\; {\rm km \cdot h^{-1}}.

Windspeed: 100\; {\rm km \cdot h^{-1}}.

Step-by-step explanation:

Assume that x\; {\rm km \cdot h^{-1}} is the speed of the plane in still air, and that y\; {\rm km \cdot h^{-1}} is the speed of the wind.

  • When the plane is travelling against wind, the ground speed of this plane (speed of the plane relative to the ground) would be (x - y)\; {\rm km \cdot h^{-1}}.
  • When this plane is travelling in the same direction as the wind, the ground speed of this plane would be (x + y)\; {\rm km \cdot h^{-1}}.

The question states that when going against the wind (v = (x - y)\; {\rm km \cdot h^{-1}},) the plane travels 6111\; {\rm km} in 9\; {\rm h}. Hence, 9\, (x - y) = 6111.

Similarly, since the plane travels 7911\; {\rm km} in 9\; {\rm h} when travelling in the same direction as the wind (v = (x + y)\; {\rm km \cdot h^{-1}},) 9\, (x + y) = 7911.

Add the two equations to eliminate y. Subtract the second equation from the first to eliminate x. Solve this system of equations for x and y: x = 779 and y = 100.

Hence, the speed of this plane in still air would be 779\; {\rm km \cdot h^{-1}}, whereas the speed of the wind would be 100\; {\rm km \cdot h^{-1}}.

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