On the first day of travel, a driver was going at a speed of 40 mph. The next day, he increased the speed to 60 mph. If he drove 2 more hours on the first day and traveled 20 more miles, find the total distance traveled in the two days.

380 miles

Step-by-step explanation:

Let the time he travelled on the second day = t

Then the first day's time is t + 2

The first day's distance is d+20

The second day's distance is d

The second day's distance is also 60*t or d = 60*t

First day: 40*(t+2)  = d + 20

Put the second day's distance into the first day's equation

40(t+ 2) = 60t + 20       Now solve for t. Remove the brackets

40*t + 80 = 60t + 20    Subtract 20 from both sides

40t + 80 - 20 = 60t      Combine

40t + 60 = 60t              Subtract 40t from both sides

60 = 60t - 40t

60 = 20t                       Divide by 20

t = 60 / 20

t = 3

Day 2s distance = 60*3 = 180

Day 1s distance = 40*5  =200

Total                                  380

You may ask where the 20 went. It is a fact relating the distance of day 1 with day 2. No one actually travelled the 20 miles. It is just that there is a difference of 20 miles in the two distances.

140 miles

Step-by-step explanation:

He drove 80 miles on the first day because 40 * 2 = 80 (40mph for 2 hours)

He then only traveled 60 miles for one hour. the answer is 140 miles

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb