A tour bus normally leaves for its destination at 5:00 p. m. for a 200 mile trip. This week however, the bus leaves at 5:40 p. m. To arrive on time, the driver drives 10 miles per hour faster than usual. What is the bus’s usual speed?

The usually speed of the bus is 50 miles/h.

Step-by-step explanation:

Let the usual speed of the bus be x mile/hour.

We know that

The bus travels 200 miles.

To reach its destination it takes time h

This week however the bus leaves at 5:40.

The bus late 40 minutes

Now the speed of the bus is = (x+10) miles/h

The new time to reach the destination is h

According to the problem,

⇒2(x²+10x)=200×10×3

⇒x²+10x = 3000

⇒x²+10x -3000=0

⇒x²+60x-50x-3000=0

⇒x(x+60)-50(x+60)=0

⇒(x+60)(x-50)=0

⇒x= -60,50

∴x=50 [since speed does not negative]

The usually speed of the bus is 50 miles/h.

-4

step-by-step explanation:

do 6-10=

if u need it to be negative.

false

step-by-step explanation:

rulers and compasses arent required in order to bisect an angle because when bisecting angles, you're making a "half circle" and considering in order to bisect an angle it wouldn't be straight, you wouldn't need either of the two