All categories

3 years ago

Question 13 of 25

Mathematics

1 answer:

geniusboy [140]3 years ago

3 0

The answer to this question is D

You might be interested in

What is the equation in point-slope form of the line that passes through the point (−1, −3) and has a slope of 4?

Nimfa-mama [501]

The equation of line in point-slope form of the line that passes through the point (−1, −3) and has a slope of 4 is:

y+3=4(x+1)

Step-by-step explanation:

Given

m=4

and

Point = (x1,y1) = (-1,-3)

The general form of point-slope equation of line is:

$y-y_1=m(x-x_1)\\Putting the values of m, x_1 and y_1\\y-(-3)=4(x-(-1))\\y+3=4(x+1)$

Hence,

The equation of line in point-slope form of the line that passes through the point (−1, −3) and has a slope of 4 is:

y+3=4(x+1)

Keywords: Point-slope form, Slope

Learn more about equation of line at:

#LearnwithBrainly

5 0

3 years ago

Help pls! I need help to finish :(<br> No Links!

atroni [7]

The answer is 7.5.
7 4/8 is also 7.5

5 0

2 years ago

The circumference of a circle is 32 cm. Which is the closest to the radius?

vitfil [10]

I believe the answer the answer is 5.1 b

5 0

3 years ago

What is eight out of eight as a fraction

Svetlanka [38]

Answer: one whole

Step-by-step explanation:

8/8 = one whole

3 0

3 years ago

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 $-1\le x$ .

6 0

3 years ago