Ask question

Ask question

All categories

Keith_Richards [23]

1 year ago

13

Find the area of the circle with the following dimension Radius = 5 in The area is : in2 (Round to the nearest tenth as needed.)

1 answer:

tigry1 [53]1 year ago

6 0

Answer:

Can u repeat the question. I didn't understand the area.

You might be interested in

Please help me out. I need help ASAP

katrin [286]

B and D and C in that order

4 0

2 years ago

Read 2 more answers

Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$

We get the $a^6b^2$ term when $k=2$ :

$k=2 \implies \dbinom 82 2^{8-2\cdot2} a^{8-2} b^2 = 28 \cdot2^4 a^6 b^2 = \boxed{448} \, a^6b^2$

6 0

8 months ago

¿Qué número NO pertenece al conjunto de los números racionales ?

ryzh [129]

Answer:

D

Step-by-step explanation:

5 0

2 years ago

Amy is solving the equation using the steps shown.

Delicious77 [7]

The correct next step Amy could take is to factor (x^2 - 16) on the left side of the equation to yield (x + 4)(x – 4)(x + 2) = 0.

The equation is given as:

$\mathbf{(x^2 - 16)(x + 2) = 0}$

The possible next steps are:

1. Expand the equation as follows

$\mathbf{x^3+2x^2-16x-32 = 0}$

2. Express x^2 -16 as a difference of two squares as follows

$\mathbf{(x + 4)(x- 4)(x + 2) = 0}$

However, the correct step in this case is to factor x^2 - 16 as a difference of two squares to give $\mathbf{(x + 4)(x- 4)(x + 2) = 0}$

This is so, because the question says Amy is trying to solve for variable x.

This step will ensure that she gets the possible values of x, unlike expanding the whole equation .

Hence, the correct next step is $\mathbf{(x + 4)(x- 4)(x + 2) = 0}$

Read more about equations at:

brainly.com/question/7674413

7 0

1 year ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

5 0

2 years ago