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ycow [4]
3 years ago
5

Help me please Thanks

Mathematics
1 answer:
Klio2033 [76]3 years ago
4 0
The answer is 1;3 because if you simplified 2;6 you get 1;3 so the answer to that is
a: 1;3
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What is the surface area of a right cylinder which has a base with radius 9 units and has a height of 12 units?
Dmitriy789 [7]

Answer:

1187.52

Step-by-step explanation:

A=2πrh+2πr2=2·π·9·12+2·π·92≈1187.52202

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3 years ago
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Dan received image off a pair of dress shoes that originally cost $54.00 and 20% off a pair of tennis shoes that originally cost
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hey preston im watching ur 7thgrade butt

Step-by-step explanation:

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3 years ago
Jake works at a fast food restaurant last week he worked 25 hours and earned 210.75 what is jakes hourly pay
umka2103 [35]

Answer:

$8.43 perhour

Step-by-step explanation:

210.75 / 25 = 8.43

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3 years ago
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
Can a translation and a reflection map TriangleQRS to TriangleTUV?
andrey2020 [161]

Answer:

Yes

Step-by-step explanation:

Yes, a translation mapping vertex Q to vertex T and a reflection across the line containing QS will map △QRS to △TUV. No, the triangles are obtuse. Yes, a translation mapping vertex S to vertex T and a reflection across the line containing RS will map △QRS to △TUV.

5 0
3 years ago
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