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7. Solve the following problems. Draw a diagram if needed.

a. Jack rode his bike 10 miles in 30 minutes. What was his approximate speed in mph?

b. Chris placed a ladder 4 ft from the base of a 12 ft wall. If the ladder reaches the top, how tall is the ladder?


Sagot :

Sure, let's break this down step-by-step.

### Part a:
Problem:
Jack rode his bike 10 miles in 30 minutes. We need to determine his speed in miles per hour (mph).

Solution:
1. Convert the time from minutes to hours.
- Given time is 30 minutes.
- Since there are 60 minutes in an hour, we convert 30 minutes to hours:
[tex]\( 30 \text{ minutes} = \frac{30}{60} \text{ hours} = 0.5 \text{ hours} \)[/tex].

2. Calculate Speed:
- We know the formula for speed is:
[tex]\[ \text{Speed} = \frac{\text{Distance}}{\text{Time}} \][/tex]
- Given distance is 10 miles and time is 0.5 hours:
[tex]\[ \text{Speed} = \frac{10 \text{ miles}}{0.5 \text{ hours}} = 20 \text{ mph} \][/tex]

Therefore, Jack's speed was 20 miles per hour (mph).

### Part b:
Problem:
Chris placed a ladder 4 feet from the base of a 12-foot wall. We need to determine the length of the ladder if it reaches the top of the wall.

Solution:
1. Visualize and use the Pythagorean theorem:
- Picture a right triangle where:
- One leg of the triangle is the distance from the base of the wall to the ladder, which is 4 feet.
- The other leg of the triangle is the height of the wall, which is 12 feet.
- The hypotenuse of the triangle is the length of the ladder.
- The Pythagorean theorem states:
[tex]\[ a^2 + b^2 = c^2 \][/tex]
where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the legs of the triangle, and [tex]\(c\)[/tex] is the hypotenuse (the ladder in this case).

2. Substitute the given values and solve for the hypotenuse (c):
- [tex]\(a = 4\)[/tex] feet (base distance)
- [tex]\(b = 12\)[/tex] feet (wall height)
- [tex]\(c =\)[/tex] length of the ladder
[tex]\[ 4^2 + 12^2 = c^2 \\ 16 + 144 = c^2 \\ 160 = c^2 \\ c = \sqrt{160} \approx 12.649110640673518 \][/tex]

Therefore, the length of the ladder is approximately 12.65 feet.
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