What is the equation of a line with slope -2 and passes through the point (0, 3)?

Mathematics

1 answer:

kramer3 years ago

5 0

The equation that you are looking for is y = -2x + 3

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Tiva earns $48 for 6 hours of babysitting. Complete each statement if Tiva keeps earning her babysitting money at this rate. CLE

lawyer [7]

8.5
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$8 \times 8.5$
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2 years ago

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An asteroid follows a hyperbolic path defined by the equation 16x2 − 9y2 = 576. If one of its foci is the sun's position, the mi

amid [387]

The minimum distance between the asteroid and the sun is 10 - 6 =4

The given hyperbolic path defined as
divide both the sides by 576,

The general equation of

hyperbola is written as
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Compare with above mention equation
Here

The distance between the asteroid and the sun is seen by figure -1
or
⇒c =10

The minimum distance between the asteroid and the sun is 10 -6 =4

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3 years ago

Continuous Probability

soldi70 [24.7K]

There would be about a 14.3% chance of being able to see it
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Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e

Bingel [31]

Assuming the series is

$\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}$

The series will converge if

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|$

So the series will certainly converge if $-1$ , but we also need to check the endpoints of the interval.

If $x=1$ , then the series is a scaled harmonic series, which we know diverges.

On the other hand, if $x=-1$ , by the alternating series test we can show that the series converges, since

$\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0$

and is strictly decreasing.

So, the interval of convergence for the series is $-1\le x$ .

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Math plzzz helppp.due today!!!!will mark brainliest

mario62 [17]

Answer:

I think that's right. I checked my answer and proved it on that sheet.

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