No idea what to do here help is appreciated :)
2 answers:
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Answer:
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.
Answer:
Step-by-step explanation:
- The concept of demand is applied, demand may be elastic, inelastic and unitary as the detailed steps and calculations is as shown in the attached file.
- For elastic demand, P > 1
- For inelastic demand, P < 1
- For unitary, P = 1
