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2 years ago

The relationship between y and x is 9/3 which table represents this relationship and why?

Mathematics

1 answer:

svetoff [14.1K]2 years ago

7 0

Answer:

First one

Step-by-step explanation:

For every 1 in x the y goes up by 3

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Answer:

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5 0

3 years ago

1/3x(-6+27y-51z)simplify

Naya [18.7K]

Final result :

9y - 17z - 2
Step by step solution :

Step 1 :

1
Simplify —
3
Equation at the end of step 1 :

1
— • (27y - 51z - 6)
3
Step 2 :

Pulling out like terms :

3.1 Pull out like factors :

27y - 51z - 6 = 3 • (9y - 17z - 2)

Final result :

9y - 17z - 2

8 0

3 years ago

Read 2 more answers

1. 14=7b 2. -11=k/4 3.6z=-6

Romashka-Z-Leto [24]

Answer:

the anwer is 14-2294-dhi2

Step-by-step explanation:

Because 14 is - 4 you multiple to 5 equal= dhi2

3 0

2 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

Can someone be my friend please!!!

KIM [24]

Answer:

no problem sure!

Step-by-step explanation:

4 0

2 years ago

The relationship between y and x is 9/3 which table represents this relationship and why￼?

The relationship between y and x is 9/3 which table represents this relationship and why?