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3 years ago

Is 3y=12 a linear function and if so, how would you graph it?

Mathematics

2 answers:

Fynjy0 [20]3 years ago

6 0

find y which is 4 (12/3) then in would be a straight line horizontally because y is going to always equal 4. Don't know if this helps but here you go. Have a good day.

alexdok [17]3 years ago

4 0

3y = 12 divide both sides by 3

y = 4

It's a linear function with a slope 0.

It's a horizontal line (look at the picture).

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Solve for x: 1/2(4x+3)=3x-5

Temka [501]

Answer:

I believe x=6 1/2 which is 6.5 in decimal form :)

8 0

3 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

What is the probability that the first card is a kingking and the second card is a kingking if the sampling is done with replac

IRINA_888 [86]

You want to draw 2 kings from a 52 card deck. And you do it with replacement.
There are 4 kings in a standard deck. The probability of getting one of them is
4/52 on the first draw.

For the second draw the probability is the same.
4/52

The probability for both happening is
(4/52)*(4/52) = (1/13)*(1/13) = 1/169 = 0.001597

5 0

3 years ago

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Darya [45]

A Jew on juice and H

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2 years ago

(picture included) brainliest

Dimas [21]

Step-by-step explanation:

118 is the answer 238-120 = 118

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3 years ago