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2 years ago

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Mathematics

1 answer:

Over [174]2 years ago

7 0

Y= 13 over 10x + 2 is the answer. The most important thing to remember when finding the slope of the line is RISE over RUN. When you rise, you go up or down. When you run, you go right or left. Think of RISE over RUN as a fraction.

You find two points on the line and you count up/down a certain amount of spaces, and right/left a certain amount of spaces to go from one point to the other point.

In this equation, we could use the point on the y-axis and the last point we see on the graph located at (10,15). You count up/down vertically first until you reach the horizontal line that the point is on. Then, you count the amount of spaces horizontally it takes you to reach the point. In this problem, you move up 13 spaces, and move to the right 10 spaces. This as a fraction will be 13 over 10, because we rose 13 spaces and we ran 10 spaces to get to our second point. The y-intercept will be the only number that is on the y-axis, which is 2.

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2 years ago

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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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