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Maksim231197 [3]
3 years ago
8

1/3 (x-10) = -4 what is x

Mathematics
2 answers:
-BARSIC- [3]3 years ago
7 0

Step-by-step explanation:

1/3x-10/3=-4

1/3x=-4+10/3

1/3x=-2/3

x=-2

adell [148]3 years ago
5 0

Answer:

Solve for x by simplifying both sides of the equation, then isolating the variable.

x = −2

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Choose the correct molecular geometry of the phosphorus atom in each of these ions from the list below: A) square plane B) T-sha
mario62 [17]

Answer:

See attached picture.

Step-by-step explanation:

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For remaining parts resubmit question.  

5 0
3 years ago
Solve for x. Enter your answer in interval notation using grouping symbols <br> X^2 + 5x &lt; 24
icang [17]
I think the correct answer would be : (-8,3)
I hope this helps !
7 0
3 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
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AnnZ [28]

Answer:

:)

Step-by-step explanation:

5/15 divided by 5 is equal to 1/3 and

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You have to divide the numerator with the same number that you divide the denominator with.  

6 0
2 years ago
PLEASE HELP! ! ! urgent!
sertanlavr [38]

Answer:

6(\frac{1}{3} \times\frac{7}{5} )=(6\times\frac{1}{3} )\frac{7}{5}

Answer: B

<u>-------------------------</u>

Hope it helps...

Have a great day!!

8 0
3 years ago
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