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Readme [11.4K]
4 years ago
15

Use the distributive property to simplify the expression 5(3 +x) – 2

Mathematics
2 answers:
Anna007 [38]4 years ago
7 0

Hello!

<em><u>Answer: 5x+13</u></em>

Explanation:

Distributive property: → a(b+c)=ab+ac

Expand the form.

5(3+x)

multiply by 5*3.

5*3=15

=15+5x

15+5x-2

Simplify by equation or expand.

15+5x-2=5x+13

Hope this helps!

insens350 [35]4 years ago
4 0

5(3 +x) – 2 = (5+3)*(5+x)

                       (8-2)*(5x-2)

                         6*5x-2

this Is what I came up with.

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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
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.020 because it is 2 spots behind the decimal

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3 years ago
Assume that 2 cards are drawn from a standard 52-card deck. Find the following probabilities.
topjm [15]

Answer:

Step-by-step explanation:

WITHOUT replacement of first card drawn:

P(a 10 is drawn) = 13/52 = 1/4

P(the next draw is a 10) = 12/52 = 3/13

P(drawing two 10s without replacement of the first draw) = (1/4)(3/13) = 3/52

WITH replacement of first card:

P(two 10s are drawn) = P(first card is a 10)*P(first card is a 10) = (4/13)(4/13) =

16/169

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Answer:

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