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2 years ago

7

There are two bowls. One of them has 60 gems and the other has 40 gems. Some of them are red. All the gems from the two bowls ar

e now put into a third bowl. What fraction of the gems in the third bowl will be red?

1 answer:

uranmaximum [27]2 years ago

5 0

Answer:

60/40

the awnser for the question is 3/2

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Which shows 522 − 482 being evaluated using the difference of perfect squares method?

Natalija [7]

52² − 48² = (2704 + 2304)(2704 − 2304) = 2,003,200

52² − 48² = (52 + 48)(52 − 48) = 100 * 4 = 400
a² - b² = (a +b)*(a-b)

52² − 48² = 2704 − 2304 = 400

<span>52</span>²<span> − 48</span>²<span> = (52 − 48)</span>²<span> = (4)</span>²<span> = 16</span>

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The cost of every college is expected to increase 3.5% next year.

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Answer:

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Step-by-step explanation:

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2 years ago

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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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2 years ago

o-na [289]

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2 years ago

How do I find these equations using substitution y=-3x+5 and 5x-4y=-3 ?

lapo4ka [179]

Answer:

x=1

Step-by-step explanation:

you would change it to 5x-4(-3x+5)=-3, because you put the info for the first equation into the second equation.

5x-4(-3x+5)=-3

5x+12x-20=-3

17x-20=-3

17x=17

x=1

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