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OLEGan [10]
2 years ago
9

What is 1+1=?apparently it's window

Mathematics
1 answer:
poizon [28]2 years ago
6 0
Possible answers:
2
11
window
ii
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NEED HELP ASAP
VLD [36.1K]

the perimeter of each side is 5 so the answer is 20

3 0
3 years ago
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If i have a 20 dollar bill and a candy is 4 how many can i buy
8090 [49]

Answer:

a candy is $4

you have a $20 bill

1 candy = 4

2 candies = 8

3 candies = 12

4 candies = 16

5 candies = 20

you can buy 5 candies for $20.

4 0
3 years ago
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I think u guys know what to do !!!
kotykmax [81]

Answer:

B, C

Step-by-step explanation:

7 - x = 18

A.

7 - 2x = 18 - x

126 - 14x = 324 - 18x

We are nowhere near the solution.

B.

-x = 11

x = -11

This works.

C.

7 = x + 18

-11 = x

This works.

D.

25 - x = 36

-25 + x = -36

This does not help.

E.

7 = x + 18

0 = x + 11

We still do not have a solution.

Answer: B, C

4 0
2 years ago
A
Semmy [17]

Step-by-step explanation:

y=75+50(36 hours)=1875

give me brainliest

4 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
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