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pishuonlain [190]
3 years ago
13

Janelle has 75$. She buys jeans that are on sale for half the price and then uses an In store coupon for 10$ off. After paying $

1.80 in sales tax, she receives $37.20 in change. What was the original price of jeans?
Mathematics
1 answer:
Nadya [2.5K]3 years ago
6 0
I think that the answer is $92
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What is x? if u can’t see it the 2 other degrees are 32 and 120
Westkost [7]

Answer:28

Step-by-step explanation:

Since x+32 is 90 degrees, you subtract 32 from both sides to get the remaining which is 28 degrees.

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What conclusions can be drawn using the given and its diagram?
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Answer:

Whichever reasoning processes and research methods were used, the final conclusion is critical, determining success or failure. If an otherwise excellent experiment is summarized by a weak conclusion, the results will not be taken seriously.

Success or failure is not a measure of whether a hypothesis is accepted or refuted, because both results still advance scientific knowledge.

Failure lies in poor experimental design, or flaws in the reasoning processes, which invalidate the results. As long as the research process is robust and well designed, then the findings are sound, and the process of drawing conclusions begins.Whichever reasoning processes and research methods were used, the final conclusion is critical, determining success or failure. If an otherwise excellent experiment is summarized by a weak conclusion, the results will not be taken seriously.

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Step-by-step explanation:

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3 years ago
Which set is a function?
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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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3 years ago
821,800 to the nearest hundred
vova2212 [387]

Answer:821,800

Step-by-step explanation:

3 0
3 years ago
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