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zhuklara [117]
3 years ago
13

How do i do this? i’m having trouble

Mathematics
1 answer:
Fantom [35]3 years ago
6 0

Answer:

121

Step-by-step explanation:

5x-54=3x+16

5x-3x=16+54

2x=70

x=35

-------

mAB= 5x-54= 5*35-54= 175-54= 121

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An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh
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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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Step-by-step explanation:

Prime factorization: 43 is prime. The exponent of prime number 43 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 43 has exactly 2 factors.

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