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2 years ago

Find the value ofx in the equation 3x+1=×+9.

Mathematics

2 answers:

fomenos2 years ago

6 0

Answer:

x = 4

Step-by-step explanation:

First, you subtract 1 from both sides, giving you 3x = x + 8

Then you subtract 1x from both sides, so 2x = 8

Divide the 2 from both sides which leaves you with the answer, x = 4

Ivanshal [37]2 years ago

6 0

Answer:

Step-by-step explanation:

3x+1=x+9

3x-x=9-1

2x=8

$\frac{2x}{2} = \frac{8}{2 }$

x=4

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Hunter-Best [27]

Answer: 11

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Hope that had helped

4 0

2 years ago

Read 2 more answers

If RSV= (5x-1), VST= (17x-6), and RST= 147, find VST

Kisachek [45]

<h3>The value of VST is 4 cm.</h3>

<h3>Hope it helps...</h3>

6 0

2 years ago

Read 2 more answers

Someone smart help me please

mafiozo [28]

Pretty sure it is
shift 1 left
i might be wrong

7 0

2 years ago

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:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

2 years ago

Verona is solving the equation –3 + 4x = 9. In order to isolate the variable term using the subtraction property of equality, wh

frutty [35]

Answer:

subtract -3

Step-by-step explanation:

–3 + 4x = 9

Add 3 to each side

This is the same as subtracting -3

-3 + 4x - (-3) = 9 - (-3)

4x = 9 +3

4x = 12

8 0

2 years ago

Read 2 more answers

Find the value ofx in the equation 3x+1=×+9.​

Find the value ofx in the equation 3x+1=×+9.