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PilotLPTM [1.2K]
3 years ago
11

What is the solution set? −4.9x+1.3>11.1 Enter your answer in the box.

Mathematics
1 answer:
Stells [14]3 years ago
8 0
Begin by adding 4.9x to both sides.
1.3> 11.1 + 4.9x This way you never have to remember to turn the > into < when dividing by a minus.

Subtract 11.1 from both sides
-9.8 < 4.9x Divide by 4.9
-9.8 / 4.9 < x
-2 < x

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What is the answer to 11
zlopas [31]

Answer:

14

Step-by-step explanation:

4x4 -2

6 0
3 years ago
Can someone help me with system of equations?
nignag [31]
2x + y = 350
x + 2y = 475 
3x + 3y = 825
x + y = 275
x = 275 - y 
(plug back into first equation) 
2(275 - y) + y = 350
550 - 2y + y = 350 
550 - y = 350
-y =-200 or y = $200 (dog)
therefore, x = $75 (cats)
Check
2(75) + 200 = 350
75 + 400 = 475 
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3 0
3 years ago
What is the value of the digit 5 in this number?<br> 60.35<br> A.0.5<br> B.0.05<br> C.5.0<br> D.5.00
Sergeeva-Olga [200]
B because its in the hundredths place
6 0
3 years ago
Multiply then write the product in the simplest form <br> 3/1
Ganezh [65]

Answer:

9

Step-by-step explanation:

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3*3= 9

Answer: 9

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(I think this is what it means)

6 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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