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In-s [12.5K]
3 years ago
9

A marathon is 26.2 miles. What is the least number of times Miguel must run for his total distance run during training to exceed

the distance of a marathon?
Mathematics
2 answers:
ziro4ka [17]3 years ago
7 0

Missing part of the question

Miguel has started training for a race. The first time he trains, he runs 0.5 mile. Each subsequent time he trains, he runs 0.2 mile farther than he did the previous time.

What is the arithmetic series that represents the total distance Miguel has run after he has trained n times?

Answer:

The least number of times Miguel must run for his total distance run during training to exceed the distance of a marathon is 17.3 miles

Step-by-step explanation:.

Given parameters

Miguel first run = 0.5 mile

Subsequent run = 0.2 mile

This question is an arithmetic progression.

We'll make use of arithmetic progression formula to solve this

Formula:.

Tn = a + (n - 1)d

Where a = first term

n = number of terms

d = common difference

In this case

a = first run = 0.5 mile

d = subsequent run = 0.2 mile

So, Tn = a + (n - 1)d become

Tn = 0.5 + (n - 1) 0.2

Tn = 0.5 + 0.2n - 0.2

Tn = 0.5 - 0.2 + 0.2n

Tn = 0.3 + 0.2n

The arithmetic series of an arithmetic progression is calculated using

Sn = ½(a + Tn) * n

By substituton, we have

Sn = ½(0.5 + 0.3 + 0.2n) * n

Sn = ½(0.8 + 0.2n) * n

Sn = 0.4n + 0.1n²

b.

Since the race is 26.2 miles then the least number of times is given as

Sn ≥ 26

0.4n + 0.1n² ≥ 26.2

0.1n² + 0.4n - 26.2 ≥ 0

Using quadratic formula

n = (-b ± √(b² - 4ac))/2a

Where b = 0.4 a = 0.1 and C = -26.2

So,

n = -0.4 ± √(0.4² - 4 * 0.1 * ,26.2)/2 * 0.1

n = (-0.4 ± √10.64)/0.2

n = (0.4 ± 3.26)/0.2

n = (0.4 + 3.26)/0.2 or (0.4 - 3.26)/0.2

n = 3.46/0.2 or -2.86/0.2

n = 17.3 or -14.3

Since n can't be negative

n = 17.3 miles

The least number of times Miguel must run for his total distance run during training to exceed the distance of a marathon is 17.3 miles

victus00 [196]3 years ago
7 0

Answer:

15 times

Step-by-step explanation:

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3 years ago
charles deposited $12,000 in the bank. He withdrew $5,000 from his account after one year. If he recives a total amount of $9,34
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Answer:

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Step-by-step explanation:

* Lets talk about the simple interest

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# A = Total amount (principal + interest)

# P = Principal amount

# I = Interest amount

# r = Rate of Interest per year in decimal r = R/100

# R = Rate of Interest per year as a percent R = r * 100

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∵ I = Prt

∵ P = 12000

∵ t = 1

∴ I = 12000(r)(1) = 12000r

- Lets subtract the money that he withdrew

∵ He withdrew $5000

∵ He deposit at first 12000

∴ He has after the withdrew 12000 - 5000 = 7000

- The new P for the next 2 years is 7000

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And he drove 50 miles for (5-X) hours, then his average rate in the afternoon was 50/(5-X) mph.

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