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Kruka [31]
3 years ago
6

HELP 15 POINTS Write a rule for the nth term of the sequence. Then find a20 - 51,48,45,42

Mathematics
1 answer:
garri49 [273]3 years ago
7 0

Answer:

45 because it is what I think

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Due to the nature of spheres, they are all similar to each other.
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What is the equation of the line that has a slope of 4 and passes through the point (3, -10?
irga5000 [103]
Hello 
the equation is y = ax+b
a the slop    a = 4
y = 4x+b
the line <span>passes through (3, -10) :  -10 = 6(3)+b
b = -28
</span>the equation is y = 6x-28
3 0
3 years ago
HELP ASAP
stira [4]

Answer:

Use khan academy for the graph.

Step-by-step explanation:

8 0
3 years ago
5 and 1/7 * 8 <br><br>HELP ME PLZ​
il63 [147K]

Answer:

41 1/7

Step-by-step explanation:

5 1/7 x 8

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36  x  8

----      --

7   x    1

288/7

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41 1/7

---

hope it helps

8 0
3 years ago
A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021
egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

  • A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

  • \displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

  • \displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}

Therefore, for the last term we have;

  • \displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

\displaystyle \frac{A}{2}  = \mathbf{ 1011 \cdot  \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460}  \right)}

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2} +\frac{1}{2} -  \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022}  \right)

Which gives;

\displaystyle \frac{A}{2}  = 1011 \cdot  \left(1 - \frac{1}{2022}  \right)

\displaystyle  A = 2 \times 1011 \cdot  \left(1 - \frac{1}{2022}  \right) = \frac{1032231}{511} \approx \mathbf{2020.022}

  • A ≈ <u>2020.022</u>

Learn more about the sum of a series here:

brainly.com/question/190295

8 0
2 years ago
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