during the first part of a trip, a conoeist travels 70 miles

at a certain speed. the canoeist travles 7 miles on the second
part of the trip at a speed of 5 mph slower. the total time
for the trip is 5 hrs. Ehat is the speed on each part of the trip?

Assume the speed for the first trip is v and the speed for the second trip is

(v - 5)

Since time = distance/speed, you get:

(70)/(v) + (7)/(v - 5) = 5 hrs.

Solve for v to get the first speed, then subtract 5 to get the second speed.

To solve this problem, we need to set up equations based on the information provided.

Let's assume the speed during the first part of the trip is 'x' mph.

1. First Part of the Trip:
Distance = 70 miles
Speed = x mph
Time = Distance / Speed = 70 / x hours

2. Second Part of the Trip:
Distance = 7 miles
Speed = (x - 5) mph (5 mph slower than the speed during the first part)
Time = Distance / Speed = 7 / (x - 5) hours

3. Total Time for the Trip:
The sum of the time for the first and second parts should be equal to 5 hours:
Time for First Part + Time for Second Part = 5
70 / x + 7 / (x - 5) = 5

Now, let's solve the equation:

Multiply all terms by x(x - 5) to get rid of the denominators:
70(x - 5) + 7x = 5x(x - 5)

Distribute and simplify:
70x - 350 + 7x = 5x² - 25x
77x - 350 = 5x² - 25x

Rearrange the equation to form a quadratic equation:
5x² - 25x - 77x + 350 = 0
5x² - 102x + 350 = 0

Factor or use the quadratic formula to solve for x:
Since the equation does not factor easily, we can use the quadratic formula:

x = [-b ± √(b² - 4ac)] / 2a

Using a = 5, b = -102, and c = 350, we can substitute these values into the formula:

x = [-(-102) ± √((-102)² - 4 * 5 * 350)] / 2 * 5

Simplifying further, we get:

x = [102 ± √(10404 - 7000)] / 10
x = [102 ± √(3404)] / 10

Now we have two possible speeds for the first part of the trip.

x₁ = [102 + √(3404)] / 10
x₂ = [102 - √(3404)] / 10

Calculating these values using a calculator:

x₁ ≈ 14.96 mph
x₂ ≈ 0.54 mph

Since the canoeist's speed cannot be negative or less than 7 mph (because it is already 5 mph slower on the second part), we conclude that the speed during the first part of the trip is approximately 14.96 mph, and the speed during the second part is approximately (14.96 - 5) mph, which is 9.96 mph (rounded to two decimal places).