Answer: The outbound trip is 18 miles per hour
Step-by-step explanation:
Let x represent the speed of the plane.
An aircraft carrier made a trip. Let us assume that this trip was outbound. The trip there took 5 hours.
Distance = speed × time. Therefore
Distance = 5x
The trip back took 6 hours. It averaged 3 mph faster on the trip there then on the return trip. This means that the speed on the trip back is x - 3 mph.
Distance = 6(x-3) = 6x - 18
Since the distance is the same,
5x = 6x - 18
6x - 5x = 18
x = 18
The speed on return or inbound trip would be 18 - 3 = 15 mph
Cot (- 210º) = - √3 so <span>30° </span>
Answer:
Step-by-step explanation:
1) 9.5 x 0.14
= 19/2 x 7/50
= 133/100
= 1.33
9.5 - 1.33
= <u><em>8.17</em></u>
<u><em></em></u>
2) 60 - 46.20 / 60
= 13.80 / 60
= 0.23
= <u><em>23% decrease</em></u>
The given system of equations is :
x−y=−11 ...(1)
y+7=−2x ...(2)
First step is to solve equation (1) for x so that we can substitute that x in equation (2) to get the value of y. So, add y to each sides of equation (1). Hence,
x = y - 11.
Next step is to plug in y - 11 for x in equation (2). So,
y + 7 = -2(y - 11)
y + 7 = -2y + 22 By distribution property.
y + 7 - 7 = -2y + 22 - 7 Subtract 22 from each sides.
y = -2y + 15
y + 2y = -2y + 15 + 2y Add 2y to each sides of the equation.
3y = 15
Divide each sides by 3.
So, y = 5.