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timofeeve [1]
3 years ago
11

When Joycelyn goes to work, she drives at an average speed of 65 miles per hour. It takes about 1 hour and 30 minutes for Joycel

yn to arrive at work. Her car travels about 25 miles per hour of gas. If gas cost $3.45 per gallon, how much money does Jocelyn spend on gas to travel to work?
Mathematics
1 answer:
insens350 [35]3 years ago
4 0

Jocelyn has to spend $13.46 on gas.

To find this, you first have to determine the number of miles she drives. You can find this by multiplying the speed by the travel time.

65 miles per hour * 1.5 hours = 97.5 miles.

Now that he have the amount of miles she drives, we can find the amount of gas she spends. In order to do that we'll divide the distance by the rate.

97.5 miles / 25 miles per gallon of gas = 3.9 gallons

Now we can determine the cost by multiplying the amount of gallons by the price.

3.9 gallons * $3.45 per gallon = $13.46

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The answer is in step-by-step explanation

Step-by-step explanation:

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agasfer [191]

Answer:

(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}

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∴ (10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}

∵ 5C1 = 5

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∵ 5C5 = 1

∴ (10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})

∴ (10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}

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