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vredina [299]
3 years ago
14

What is the independent variable of y=12x−2

Mathematics
1 answer:
alexdok [17]3 years ago
3 0
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What is the cos of 28? I need help :( please
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I hope this helps you


COS28=15/17
5 0
3 years ago
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What is 6 to the 2nd power ÷ 2(3)+4
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6^2:2(3)+4=36:2(3)+4=18(3)+4=54+4=\huge\boxed{58}
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3 years ago
Find the median of this data: 18, 9, 17, 18, 15, 10, 18. If necessary, round to the nearest tenth.
Marta_Voda [28]
9, 10, 15, 17, 18, 18, 18

17

3 0
2 years ago
The sequence$$1,2,1,2,2,1,2,2,2,1,2,2,2,2,1,2,2,2,2,2,1,2,\dots$$consists of $1$'s separated by blocks of $2$'s with $n$ $2$'s i
kicyunya [14]

Consider the lengths of consecutive 1-2 blocks.

block 1 - 1, 2 - length 2

block 2 - 1, 2, 2 - length 3

block 3 - 1, 2, 2, 2 - length 4

block 4 - 1, 2, 2, 2, 2 - length 5

and so on.


Recall the formula for the sum of consecutive positive integers,

\displaystyle \sum_{i=1}^j i = 1 + 2 + 3 + \cdots + j = \frac{j(j+1)}2 \implies \sum_{i=2}^j = \frac{j(j+1) - 2}2

Now,

1234 = \dfrac{j(j+1)-2}2 \implies 2470 = j(j+1) \implies j\approx49.2016

which means that the 1234th term in the sequence occurs somewhere about 1/5 of the way through the 49th 1-2 block.

In the first 48 blocks, the sequence contains 48 copies of 1 and 1 + 2 + 3 + ... + 47 copies of 2, hence they make up a total of

\displaystyle \sum_{i=1}^48 1 + \sum_{i=1}^{48} i = 48+\frac{48(48+1)}2 = 1224

numbers, and their sum is

\displaystyle \sum_{i=1}^{48} 1 + \sum_{i=1}^{48} 2i = 48 + 48(48+1) = 48\times50 = 2400

This leaves us with the contribution of the first 10 terms in the 49th block, which consist of one 1 and nine 2s with a sum of 1+9\times2=19.

So, the sum of the first 1234 terms in the sequence is 2419.

8 0
1 year ago
What is 3/18 divided by 7/21... help plz
saw5 [17]
3/18 divided 7/21 =3/18 multiply by 21/7 (we have to flip the second fraction and multiply by 3/18)= 63/126=1/2
So the answer is 1/2
8 0
3 years ago
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