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Murrr4er [49]
2 years ago
11

Solve 5-2x > 4, then represent on a number line​

Mathematics
1 answer:
Varvara68 [4.7K]2 years ago
7 0

Answer:  x < 1/2 or x < 0.5

The graph is below

=========================================================

There are two ways to solve this

Method 1

5 - 2x > 4

-2x > 4-5

-2x > -1

x < -1/(-2) .... note the inequality sign flips

x < 1/2

x < 0.5

The inequality sign flips whenever we divide both sides by a negative number.

----------------------

Method 2

5 - 2x > 4

5 - 2x+2x > 4+2x

5 > 4+2x

4+2x < 5

2x < 5-4

2x < 1

x < 1/2

x < 0.5

----------------------

To graph this on a number line, we'll plot an open hole at 1/2 or 0.5 on the number line. Then we shade to the left to represent all values smaller than 1/2 = 0.5

The open hole signals that 0.5 itself is NOT part of the solution set.

See the diagram below.

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4 0
3 years ago
WILLGIVEBRAINLIEST,THANKS,5STARS!<br> PLEASE HELP!<br> PLEASE EXPLAIN IN DEPTH!
bearhunter [10]
<h3>Answer:</h3>

x=2

<h3>Solution:</h3>
  • In order to isolate x, we should first of all take the square root of both sides.
  • If we take the square root of the left-hand side, we will get
  • x-\displaystyle\frac{1}{2}
  • How about the right-hand side? Well, we should take the square root of the numerator (9) and the denominator (4)
  • So we have
  • x-\displaystyle\frac{1}{2} =\frac{3}{2}}
  • Move -1/2 to the right:
  • x=-\displaystyle\frac{3}{2} +\frac{1}{2}
  • x=\displaystyle\frac{4}{2}
  • x=2

Hope it helps.

Do comment if you have any query.

6 0
2 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
Find the least number which exactly divides 35,56 and 91 leaving reminder 7 in each case
Rufina [12.5K]

Answer:

The smallest number that when divided by 35;56;91leaves a remainder 7 in each cases 3640+7 equals to 3647. HERE YOUR ANSWER

3 0
3 years ago
Read 2 more answers
What is the image point of (-4,2) after a translation left 5 units and down 3<br> units?
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Answer: (-9,-1)

Step-by-step explanation:

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