Dale drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 7 hours. when dale drove home, there was no traffic and the trip only took 5 hours. if his average rate was 18 miles per hour faster on the trip home, how far away does dale live from the mountains? do not do any rounding.
Answer:
Dale live 315 miles from the mountains
Step-by-step explanation:
Let y be the speed of Dale to the mountains
Time taken by Dale to the mountains=7 hrs
Therefore distance covered by dale to the mountain = speed × time = 7y ......eqn 1
Time taken by Dale back home = 5hours
Since it speed increased by 18 miles per hour back home it speed = y+18
So distance traveled home =speed × time = (y+18)5 ...... eqn 2
Since distance cover is same in both the eqn 1 and eqn 2.
Eqn 1 = eqn 2
7y = (y+18)5
7y = 5y + 90
7y - 5y = 90 (collection like terms)
2y = 90
Y = 45
Substitute for y in eqn 1 to get distance away from mountain
= 7y eqn 1
= 7×45
= 315 miles.
∴ Dale leave 315 miles from the mountains
The answer is -4 + 2 + -0.25 + 0.5.
in form of multiplication is 
.
<u>Step-by-step explanation:</u>
We know that factors are the numbers that are being multiplied together . As 6532*7 = 45724 .Here 6532 & 7 are the factors and 45724 is product of the factors .
Here , we need to convert to the product of multiplication
a^2-b^2-4b-4 or , :
⇒
We know by identity that , 
⇒ 
Taking common term out as -4 :
⇒
Therefore , in form of multiplication is .
Answer: It is not a Function
Step-by-step explanation: One input has two outputs.
