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Semenov [28]
3 years ago
7

Describe the location of the point having the following coordinates.

Mathematics
1 answer:
Maslowich3 years ago
6 0

Answer:

Step-by-step explanation:

Quadrant II or Quadrant IV

Quadrant I or Quadrant II

Quadrant I or Quadrant IV

Quadrant III or Quadrant IV

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Allie rode her bike up a hill at an average speed of 12 feet/second. She then rode back down the hill at an average speed of 60
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Answer:

The total distance Allie traveled was 0.81 miles.

Step-by-step explanation:

Since Allie rode her bike up a hill at an average speed of 12 feet / second, and she then rode back down the hill at an average speed of 60 feet / second, and the entire trip took her 2 minutes, to determine what is the total distance she traveled, the following calculation must be performed:

12 + 60 = 72

72 x 60 = 4320

1000 feet = 0.189394 miles

4320 feet = 0.8181818 miles

Therefore, the total distance Allie traveled was 0.81 miles.

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Answer:

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Step-by-step explanation:

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family of solutions of the second-order DE y y 0. Find a solution of the second-order IVP consisting of this differential equati
Anna35 [415]

Answer:

y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}

Step-by-step explanation:

Given

y=c_1e^x +c_2e^{-x

y(1) = 0

y'(1) =e

Required

The solution

Differentiate y=c_1e^x +c_2e^{-x

y' = c_1e^x - c_2e^{-x}

Next, we solve for c1 and c2

y(1) = 0 implies that; x = 1 and y = 0

So, we have:

y=c_1e^x +c_2e^{-x

0 = c_1 * e^1 + c_2 * e^{-1}

0 = c_1 e + \frac{1}{e}c_2 --- (1)

y'(1) =e implies that: x = 1 and y' = e

So, we have:

y' = c_1e^x - c_2e^{-x}

e = c_1 * e^1 - c_2 * e^{-1}

e = c_1 e - \frac{1}{e}c_2 --- (2)

Add (1) and (2)

0 + e = c_1e + c_1e + \frac{1}{e}c_2 - \frac{1}{e}c_2

e = 2c_1e

Divide both sided by e

1 = 2c_1

Divide both sides by 2

c_1 = \frac{1}{2}

Substitute c_1 = \frac{1}{2} in 0 = c_1 e + \frac{1}{e}c_2

0 = \frac{1}{2} e+ \frac{1}{e}c_2

Rewrite as:

\frac{1}{e}c_2 = -\frac{1}{2} e

Multiply both sides by e

c_2 = -\frac{1}{2} e^2

So, we have:

y=c_1e^x +c_2e^{-x

y = \frac{1}{2}e^x -\frac{1}{2} e^2 * e^{-x}

y = \frac{1}{2}e^x -\frac{1}{2} e^{2-x}

6 0
3 years ago
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