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dusya [7]
3 years ago
8

2.943 divided by 2.7

Mathematics
1 answer:
kenny6666 [7]3 years ago
5 0

Answer:

1.09

Step-by-step explanation:

hope this helps

-mercury

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30+6⋅7+8 need help asap!!!!
Sladkaya [172]

Answer:

80 :D!

Step-by-step explanation:

30 + (6 x 7) + 8

30 + 42 + 8

72 + 8

= 80

Hope this helps! Brainliest plz?

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2 years ago
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Help with both questions please
Igoryamba
For the first question the answer is the second option.
For the second question the answer is the first option.

Hope I didn't mess up for your sake
5 0
3 years ago
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Eight friends ate ⅝ of a bag of chips. What fraction of the bag did each person eat, assuming that they each ate the same amount
Nataliya [291]

Answer:

Option B.  of the bag.

Step-by-step explanation:

Eight friends ate  of a bag of chips in the same amount.

We have to find the quantity that each person ate the chips.

Each friend ate =  ÷ 8

                         =  ×  

                         =  

Therefore, each friend ate  of the bag of chips.

Option B is the answer.

7 0
3 years ago
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54,36,24 whats the next term in the geometric sequence
kodGreya [7K]

Answer:

16

Step-by-step explanation:

The common ratio is 36/54 = 2/3.

So the next term is 2/3 (24) = 16.

3 0
3 years ago
Find the smallest positive $n$ such that \begin{align*} n &\equiv 3 \pmod{4}, \\ n &\equiv 2 \pmod{5}, \\ n &\equiv
Alex777 [14]

4, 5, and 7 are mutually coprime, so you can use the Chinese remainder theorem right away.

We construct a number x such that taking it mod 4, 5, and 7 leaves the desired remainders:

x=3\cdot5\cdot7+4\cdot2\cdot7+4\cdot5\cdot6

  • Taken mod 4, the last two terms vanish and we have

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod4

so we multiply the first term by 3.

  • Taken mod 5, the first and last terms vanish and we have

x\equiv4\cdot2\cdot7\equiv51\equiv1\pmod5

so we multiply the second term by 2.

  • Taken mod 7, the first two terms vanish and we have

x\equiv4\cdot5\cdot6\equiv120\equiv1\pmod7

so we multiply the last term by 7.

Now,

x=3^2\cdot5\cdot7+4\cdot2^2\cdot7+4\cdot5\cdot6^2=1147

By the CRT, the system of congruences has a general solution

n\equiv1147\pmod{4\cdot5\cdot7}\implies\boxed{n\equiv27\pmod{140}}

or all integers 27+140k, k\in\mathbb Z, the least (and positive) of which is 27.

3 0
3 years ago
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