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snow_lady [41]
4 years ago
13

Jessamine wants to ride her bike at least 85 miles per week. She rides to and from school 5 days per week. Her house is 6.25 mil

es from her school. She rides an additional 2.5 miles for each trip, t, she makes around the park. Which inequality represents the situation?
5 (2) (6.25) + 2.5 t less-than-or-equal-to 85
5 (2) (6.25) + 2.5 t less-than 85
5 (2) (6.25) + 2.5 t greater-than-or-equal-to 85
5 (2) (6.25) + 2.5 t greater-than 85
Mathematics
2 answers:
katrin [286]4 years ago
7 0

Answer:

the answer would be 5 (2) (6.25) + 2.5 t greater than or equal to 85

Step-by-step explanation:

Jessamine wants to ride at least 85 miles a week, so we can cross out the top 2. the 4th option is saying great than 85, but she wants to go 85 miles or greater.

aalyn [17]4 years ago
4 0

Step-by-step explanation:

Let's translate the verbal language to algebraic language.

She rides to and from school 5 days per week, 6.25 miles each route => 5*2*6.25 miles = 62.5 miles.

She rides additioanly around the park 2.5 miles for each trip t => 2.5*t = 2.5t

Total miles of her rides per week: 62.5 miles + 2.5t

She wants to ride minimum 85 miles => 62.5 + 2.5t ≥ 85

Then, the situation is represented by this inequality:

2.5t + 62.5 ≥ 85

You can develop it and get to several equivalent inequalities, for example:

2.5t ≥ 85 - 62.5

2.5t ≥ 22.5

t ≥ 9

Any of the four forms are equivalent and a valid answer.

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Step-by-step explanation:

For this case we have the following function given:

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For this case we just need to take the derivate of the position function respect to t like this:

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And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

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And for this case we got t =1s, t=5s

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

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So then the intervals positive are [0,1) \cup (5,\infty)

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

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a(t) = \frac{dv}{dt}= 6t -18

Part g and h

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And would be slowing down from [0,1) \cup (3,5)

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