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ra1l [238]
3 years ago
14

What is the nth term rule of the linear sequence below? 6, 13, 20, 27, 34, . . .

Mathematics
1 answer:
nignag [31]3 years ago
3 0

Answer:

7n - 1

Step-by-step explanation:

notice that the difference between adjacent terms is 7.

so nth term is given by:

nth term = first term + (n-1)×difference (n is any number)

= 6 + (n-1)7

= 6+ 7n -7

= 7n - 1

hope this helps :-))

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Answer would be


V= - uf/f-u


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6 0
3 years ago
2root3 sin^<a href="/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="ac9eec">[email&#160;protected]</a> - <a href
Phantasy [73]

Answer:

Step-by-step explanation:

2\sqrt{3} sin^{2} \alpha -cos\alpha =0\\2\sqrt{3} (1-cos ^2 \alpha )-cos \alpha =0\\2\sqrt{3} -2\sqrt{3} cos^2 \alpha -cos \alpha =0\\2\sqrt{3} cos^2 \alpha +cos \alpha -2\sqrt{3} =0\\cos \alpha =\frac{-1 \pm\sqrt{1^2-4*2\sqrt{3}*(-2\sqrt{3})  } }{2*2\sqrt{3} } \\=\frac{-1 \pm\sqrt{1+48} }{4\sqrt{3} } \\=\frac{-1\pm7}{4\sqrt{3} } \\either~cos \alpha =\frac{6}{4\sqrt{3} }=\frac{\sqrt{3} }{2} \\=cos \frac{\pi }{6} ,cos(2\pi -\frac{\pi }{6} )\\=cos \frac{\pi}{6} ,cos \frac{11\pi }{6}

\alpha =2 n\pi+ \frac{\pi }{6} ,2n\pi +\frac{11\pi }{6} (general~solution)

or~cos\alpha =-\frac{7}{4\sqrt{3} } \\ \alpha =cos^{-1}( \frac{-7}{4\sqrt{3} } )

5 0
3 years ago
Which steps below are in the correct order to create a congruent segment?
Alla [95]

Answer: A

Step-by-step explanation:

Thank u

8 0
3 years ago
Draw a model to represent the following problem 42 divide 7 Answer
hodyreva [135]

We have to draw the model for 42 divide 7 . In order to draw the model we use below steps.

  • Draw 7 boxes to represent 7 groups.
  • Assume 42 as the number which represents the total of these 7 groups.
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Attached is the model to represents 42 divide 7 .

In the question mark we should put 6. Hence, when we add 6 to 7 times that will be equal to 42.


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3 years ago
Help answer this question​
lyudmila [28]

Answer:

the answer should be 1/10 which is the first dash next to the zero

Step-by-step explanation:

5 0
3 years ago
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