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2 years ago

Find the area of a circle is with a diameter of 6

Mathematics

2 answers:

m_a_m_a [10]2 years ago

5 0

Answer:

28.27

Step-by-step explanation:

Mice21 [21]2 years ago

5 0

Answer:

$9\pi$ or 28.26

Step-by-step explanation:

First, you need to find the radius (6/2 =3).

Next, you need to plug the value into the equation A = pi * r^2

A = $\pi$ * 3^2

A = 9 $\pi$ = 28.26

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A researcher wishes to find out whether giving a student $1 million in cash will improve or ruin the student's grades in school.

svetoff [14.1K]

Answer:

Greta, Betty, Tuan and Hiroshi

Step-by-step explanation:

Start from the beginning:

3 - Greta

Then, continue with the next numbers:

9 - skip (there is no student numbered with a 9)

6 - Betty

6 - skip (Betty is already chosen)

1 - Tuan

5 - Hiroshi

8 0

2 years ago

The rectangle below has an area of x^2+8x+15x

Darina [25.2K]

Answer:

The length of the rectangle is x+5.

Step-by-step explanation:

Given : The rectangle below has an area of $x^2+8x+15$ square meter and a width of $x+3$ meter.

To find : What expression represents the length of the rectangle?

Solution :

The area of the rectangle is given by,

$\text{Area}=\text{Length}\times \text{Breadth}$

$\text{Area}=x^2+8x+15$

$\text{Breadth}=x+3$

$x^2+8x+15=\text{Length}\times(x+3)$

$\text{Length}=\frac{x^2+8x+15}{x+3}$

$\text{Length}=\frac{(x+5)(x+3)}{x+3}$

$\text{Length}=x+5$

Therefore, the length of the rectangle is x+5.

7 0

2 years ago

24a-22=-4(1-6a) solve for a

Naily [24]

Answer:

The srarement is false foor any value

Step-by-step explanation:

6 0

2 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

2 years ago

What is the first step to Solving Word Problems?

asambeis [7]

The first step in solving any word problem is to read the problem carefully.

Steps would be as follows:
1- Read the problem carefully
2- Identify what you are trying to find
3- Make a plan
4- Solve the problem
5- Check that your answer makes sense

8 0

2 years ago