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3 years ago

Please help idk what I’m doing

Mathematics

1 answer:

dsp733 years ago

8 0

Answer:

m∠FXM = 24°

Step-by-step explanation:

This problem has a lot of extra data that has no relation to the question. You can basically ignore any business about isosceles triangles and congruent segments.

The only thing you need to know is the angle sum theorem:

m∠FXM + m∠MXA = m∠FXA

Filling in the numbers, you get ...

m∠FXM + 72° = 96°

m∠FXM = 96° -72° = 24°

The measure of angle FXM is 24°. (matches choice A)

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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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