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zalisa
4 days ago
11

Kathleen and Arnob both run from the park entrance along a loop. Kathleen starts walking from the park entrance and gets a 5-mil

e head start on Arnob. The graph shows how far they have both traveled.
How many minutes does it take Arnob to catch up to Kathleen?

a. 9.8
b. 10
c. 73.5
d. 75

Mathematics
2 answers:
babunello [3.6K]4 days ago
5 0
<span>To tackle the above dilemma,

we utilize the formula speed = distance/time.

Observing the red line:/>A covers 8 miles in 60 minutes, thus his speed reaches 8 miles per hour while K manages 2 miles in 30 minutes, resulting in her speed of 4 miles per hour, with a 5-mile head start.</span><span>A's distance = initial gate distance + (speed x time)
A's distance = 0 + (8 x time)

K's distance = initial gate distance + (speed x time)
K's distance = 5 + (4 x time)

For them to have traveled equal distances from the gate, they must have caught up with each other.
Therefore, when:

A's distance = K's distance
0 + (8 x time) = 5 + (4 x time)
8t = 5 + 4t
8t - 4t = 5
4t = 5
t = 5/4 hours
t = 1 hour and 15 minutes
t = 75 minutes
</span>
The solution is option D. I hope this clarifies the issue.
zzz [4K]4 days ago
5 0

In this scenario, the first step is to assign variables.

We define:

x: time measured in minutes

y: distance covered.

The equations formulated are as follows:

For Kathleen:

y = (\frac{1}{15}) x + 5

For Arnob:

y = (\frac{2}{15}) x

At the moment Arnob meets Kathleen, we have:

(\frac{1}{15}) x + 5 = (\frac{2}{15}) x

Next, we isolate x.

We now have:

(\frac{2}{15}) x - (\frac{1}{15}) x = 5

(\frac{1}{15}) x = 5\\x = (15) * (5)\\x = 75 min

Answer:

Arnob takes a total of 75 minutes to catch up to Kathleen:

d. 75

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