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IgorLugansk [536]
3 years ago
13

This problem is realy tricky. Ive tried numerous ways. Can I get help?

Mathematics
1 answer:
Fiesta28 [93]3 years ago
6 0
I'm working through this problem, but I'm not getting an answer that is one of the multiple choices.  Maybe there might be a typo, or maybe there is something I don't understand.  But anyway here is my work:

First, we need to find the total hours worked, which if you add the four numbers, the total hours worked is 50.

Second, we know that they put 20% of the $1250 they earned in a savings account.  20% of $1250 is $250.  So they have $1000 left over to split amongst themselves.

One way to continue from this point is to divide the $1000 by the 50 hours to see how much money someone would make per hour.  So, a $1000 divided by 50 hours gives us $20 per hour.

In other words, for every hour that a person contributed to the landscaping business, you would earn $20.

Since Katie contributed 15.25 hours of work, she should get a total of:

15.25 x $20 = $305

Which unfortunately is not one of the answers.  I don't think my calculations are off.  But I hope that helps anyways!
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Convert 7500cm to meters
olya-2409 [2.1K]

Answer: 75 meters

Step-by-step explanation: In this problem, we're asked to convert 7,500 centimeters into meters which means we use the conversion factor for centimeters and meters.

Conversion factor ⇒ <em>100 centimeters = 1 meter</em>

Next, notice that we're converting from a smaller unit "centimeters" to a larger unit "meters" and when we convert from a smaller unit to a larger unit we divide by the conversion factor.

So we divide 7,500 by the conversion factor which is 100.

7,500 ÷ 100 = 75

Therefore, 7,500 centimeters = 75 meters.

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2 years ago
2,99,790,000 in scientific notation
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2 years ago
Determine the period
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2 years ago
Lim (n/3n-1)^(n-1)<br> n<br> →<br> ∞
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, <em>e</em> :

\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/(<em>n</em> - 9) with 1/<em>m</em>, so that

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

From the relation 9<em>m</em> = <em>n</em> - 9, we see that <em>m</em> also approaches infinity as <em>n</em> 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}

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2 years ago
Which number forms a Pythagorean Triple with 8 and 10? <br>A: 6 <br>B: 15
daser333 [38]
Are 6 and 15 the legs of the triangle?
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