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3 years ago

Rewrite the following equation in slope-intercept form.

Mathematics

2 answers:

Serga [27]3 years ago

8 0

Answer:

y = 2 $\frac{2}{5}$ x - 3 $\frac{3}{5}$

Step-by-step explanation:

Slope intercept form: y = mx + b

We need to get the equation 12x - 5y = 18 into slope-intercept form. This means, we must essentially solve for y.

12x - 5y = 18

Subtract 12x from both sides.

-5y = 18 - 12x

Divide by -5 on both sides

y = $\frac{18 - 12x}{-5}$

y = -3.6 + 2.4x

y = 2.4x - 3.6

y = 2 $\frac{2}{5}$ x - 3 $\frac{3}{5}$

We are now in slope-intercept form.

ANEK [815]3 years ago

7 0

Answer:

Step-by-step explanation:

12x - 5y = 18

-5y = -12x + 18

$y=\frac{-12}{-5}x+\frac{18}{-5}\\\\y=\frac{12}{5}x-\frac{18}{5}$

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3 years ago

An airplane travels 6111 kilometers against the wind in 9 hours and 7911 kilometers with the wind in the same amount of time. Wh

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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