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muminat
3 years ago
11

Can someone explain I’m confused ‍♂️

Mathematics
1 answer:
coldgirl [10]3 years ago
4 0

Answer:

x = 2000

y = 30000

y-intercept = 30000

slope = 15

one point = (2000 , 30000)

that is as far as I got. Sorry that I couldn't help more. :C

Step-by-step explanation:


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Help me answer for 11 points
Tcecarenko [31]

Answer:

Just Transposition...

Step-by-step explanation:

12×7=84=d

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28+57=85=f

96÷16=6=g

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95+5-29=100-29=71=I

55-6+21=70=j

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hope it helps

4 0
3 years ago
HELP DUE IN 10 MINS!<br><br> x =??
Black_prince [1.1K]

Answer:

x = 11

Step-by-step explanation:

Using Secants ad Segments Theorem we can say;

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Hope this helps!

3 0
3 years ago
Read 2 more answers
Test the series for convergence or divergence (using ratio test)​
Triss [41]

Answer:

    \lim_{n \to \infty} U_n =0

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given series is an alternating series

∑(-1)^{n} \frac{n^{2} }{n^{3}+3 }

Let   U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }

By using Leibnitz's rule

   U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }

 U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}

Uₙ-Uₙ₋₁ <0

<u><em>Step(ii):-</em></u>

    \lim_{n \to \infty} U_n =  \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }

                       =  \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }

                    = =  \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }

                       =\frac{1}{infinite }

                     =0

    \lim_{n \to \infty} U_n =0

∴ Given series is converges

                       

                     

 

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