3 years ago

SOMEONE PLEASE HELP!!! I AM IN 3RD GRADE

Mathematics

1 answer:

Evgen [1.6K]3 years ago

5 0

The answer is 1/3. I hope I’m not late but I just saw this

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The sum of a number and twice its square is 105. Find the number

IRISSAK [1]

N+2^2=105
105-4=101
n=101
hope it helps please mark me as brainliest

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2 years ago

Find the value of 4^2 + 6 ÷ 2

lorasvet [3.4K]

Answer:

Step-by-step explanation:

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4^2 + 6 ÷ 2

16 + 6 ÷ 2

16 + 3

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3 years ago

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A ___ equation is a second-degree equation of the form ax^2+bx+c=0 where a ≠ 0

klasskru [66]

Answer:

A quadratic equation is second degree equation with the standard form ax^2+bx+c=0 as the highest power is 2.

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2 years ago

WILL GIVE BRAINLIEST!!!

Charra [1.4K]

I think the answer will be A

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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