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sashaice [31]
3 years ago
7

What is the sum of the first 20 terms of the arithmetic sequence shown? \displaystyle \frac{1}{3} 3 1 ​ , \displaystyle \frac{2}

{3} 3 2 ​ , 1, \displaystyle \frac{4}{3} 3 4 ​ , \displaystyle \frac{5}{3} 3 5 ​ , ...
Mathematics
1 answer:
Murrr4er [49]3 years ago
5 0

Answer:

70

Step-by-step explanation:

a_1 = First term = \dfrac{1}{3}

d = Common difference = \dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}

n = Number of terms = 20

Sum of arithmetic progression is given by

S=\dfrac{n}{2}[2a_1+(n-1)d]\\\Rightarrow S=\dfrac{20}{2}\times (2\times \dfrac{1}{3}+(20-1)\dfrac{1}{3})\\\Rightarrow S=70

The sum of the first 20 terms of the arithmetic sequence is 70.

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yulyashka [42]
n points would divide a given line into n+1 segments.

Just assume that you have a line and you added a point to that line, this point would divide your line into two segments.
If you add two points to a line, these two points would divide your line into three segments.
So the number of segments is always greater that the number of points by 1.
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3 years ago
Help subtract (xsquared +43x10)-(10xsquared-5x+10 express the answer in standard form
ad-work [718]

Answer:

\large\boxed{(x^2+43x+10)-(10x^2-5x+10)=-9x^2+48x}

Step-by-step explanation:

(x^2+43x+10)-(10x^2-5x+10)\\\\=x^2+43x+10-10x^2-(-5x)-10\\\\=x^2+43x+10-10x^2+5x-10\qquad\text{combine like terms}\\\\=(x^2-10x^2)+(43x+5x)+(10-10)\\\\=-9x^2+48x

6 0
3 years ago
Dean needs $150 for a miming class. He already has $85 and he can earn the rest by working 13 hours at his job. If h represents
AnnyKZ [126]

Answer:

h=$65/13

Step-by-step explanation:

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So the equation is

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3 0
3 years ago
Three fair dice are rolled, one red, one green and one blue. What is the probability that the upturned faces of the three dice a
r-ruslan [8.4K]

Answer:   \dfrac{5}{9}

Step-by-step explanation:

When we throw a die , Total outcomes =6

When we throw 3 dice , Total outcomes = 6 x 6 x 6 = 216 [by fundamental counting principle]

Given : Three fair dice are rolled, one red, one green and one blue.

Favorable outcomes : When the upturned faces of the three dice are all of different numbers i.e. no repetition of numbers allowed

By Permutations , the number of favorable outcomes = ^6P_3=\dfrac{6!}{(6-3)!}=\dfrac{6!}{3!}=6\times5\times4=120

The probability that the upturned faces of the three dice are all of different numbers = \dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}

=\dfrac{120}{216}=\dfrac{5}{9}

The probability that the upturned faces of the three dice are all of different numbers  is \dfrac{5}{9} .

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3 years ago
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aliina [53]
I guess that is 1:10
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