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2 years ago

2(4 plus 2x) > 5x plus 5

Mathematics

2 answers:

egoroff_w [7]2 years ago

4 0

Answer:

x ≤3

Step-by-step explanation:

blondinia [14]2 years ago

3 0

Answer:

x ≤3

Step-by-step explanation:

2(4+ 2x)≥ 5x+ 5

distribute

8+4x ≥ 5x+ 5

Subtract 4x from each side

8+4x-4x≥ 5x-4x+ 5

8 ≥ x+ 5

Subtract 5 from each side

8-5 ≥ x+ 5-5

3 ≥ x

x ≤3

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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

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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1 year ago

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drek231 [11]

Answer:

Step-by-step explanation:

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