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Lana71 [14]
4 years ago
10

" alt="42 \times 42 \times 29 = " align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Blababa [14]4 years ago
6 0
The correct answer is 51156



Hope this helps :)
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Could someone please help I will give brainiest to best answer but please show steps?
pochemuha
It would be 9x+15

g is right above h

If two lines are cut by a transversal and the consecutive exterior angles are supplementary, then the two lines are parallel. Keep in mind you do not need to check every one of these 12 supplementary angles. Just checking any one of them proves the two lines are parallel
7 0
3 years ago
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Translate the expression into a word phrase. k + 12 A. a number less than 12 B. 12 more than a number C. the product of a number
Burka [1]
The answer is b) 12 more than a number

when you use the word "more than" it means addition. when you use the word "less than" it means subtraction. when you use the word "product of" it means multiplication. when you use "quotient of" it mean division.

hope that helps, God bless! <span />
8 0
4 years ago
The school production of​ 'Our Town' was a big success. For opening​ night, 315 tickets were sold. Students paid ​$1.50 ​each, w
Westkost [7]

Step-by-step explanation:

700.50√315=2.65

2.65×1.50=

8 0
4 years ago
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

                       

                     

 

3 0
3 years ago
I need help with this question ASAP some plz help me
madam [21]

cube root on both sides then subtract 7

answer would be (cube rooted)A-7=Z if that makes sense

4 0
3 years ago
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