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3 years ago

In the figure below, there are three right trangles. Complete the following.

Mathematics

1 answer:

lana [24]3 years ago

5 0

Answer:

Step-by-step explanation:

nis fn m

v vm

vrdac

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The Wilsons have triplets and another child who is ten years old. The sum of the ages of their children is 37. How old are the t

sesenic [268]

Hope this helps have a nice day!!!!

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3 years ago

Read 2 more answers

The food stand at the zoo sold 2,514 pounds of hamburger last month. The average cost a pound of hamburger $2. Jeremy estimates

sergiy2304 [10]

That is about right... 2,514 X 2 = 5,028 so he is about 1,000 off. I will leave the rest to you.

7 0

4 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

3 years ago

2log(base 5)^(x-3)-log(base 5)^8= log(base 5)^2

Alja [10]

Answer:

x = 7

Step-by-step explanation:

$2log_5(x-3)-log_58= log_52\\\\2log_5(x-3) =log_52+log_58\\\\2log_5(x-3) =log_5(2\times 8) \\\\2log_5(x-3) =log_516\\\\2log_5(x-3) =log_54^2 \\\\2log_5(x-3) =2 log_54 \\\\log_5(x-3) =log_54 \\\\x - 3 = 4\\\\x = 4+3\\\\\huge \orange { \boxed{x = 7}}$

7 0

3 years ago

What is 13.038 rounded to the nearest thousandth?

denis23 [38]

13.1......... I believe this is the correct answer

5 0

3 years ago