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2 years ago

The equation of a circle is x2 + (y – 10)2 = 16. The radius of the circle is units. The center of the circle is at .

Mathematics

2 answers:

Leona [35]2 years ago

8 0

Answer:

1. 4

2. (0,10)

DENIUS [597]2 years ago

4 0

Answer:

(0, 10)

Step-by-step explanation:

Compare:

x2 + (y – 10)2 = 16 and

(x - h)² + (y - k)² = 4²

The center of the circle is at (h, k), which in this case is (0, 10).

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3 years ago

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3 years ago

9. Write an equation for the table in slope-

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2 years ago

Plutonium–238 has a yearly decay constant of 7.9 × 10-3. If an original sample has a mass of 15 grams, how long will it take to

eduard

The answer is 28 years

At = A0 * e^(-k * t)
At = 12 g
A0 = 15 g
k = 7.9 × 10^-3 = 0.0079
t = ?

12 = 15 * e^(-0.0079 * t)
12/15 = e^(-0.0079 * t)
0.8 = e^(-0.0079 * t)

Logarithm both sides (because ln(e) = 1:
ln(0.8) = ln(e^(-0.0079 * t))
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2 years ago

Inverse laplace of [(1/s^2)-(48/s^5)]

Katen [24]

**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform $F(s)$ and $G(s)$ , then $\mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}$

Given that $\dfrac{1}{s^2} = \dfrac{1!}{s^2}$ and $-\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}$

From Table of Laplace Transform, you have $\mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}}$ and hence $\mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n$

So you have $\mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}$ .

Hope this helps...

7 0

3 years ago