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artcher [175]
2 years ago
9

How many sets of two or more consecutive positive integers can be added to obtain a sum of 1800?

Mathematics
1 answer:
Nataliya [291]2 years ago
4 0

Answer:

n = 60

Step-by-step explanation:

GIVEN DATA:

Total sum of consecutive number is 1800

sum of n number is given as

sum = \frac{ n(n+1)}{2}

where n is positive number and belong to natural number i.e 1,2,3,4,...

from the data given we have1800 = \frac{n(n+1)}{2}

solving for n we get

[/tex]n^2 + n -3600 = 0[/tex]

n = 59.5, -60.5

therefore n = 60

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A snowstorm began on Monday evening. It snowed steadily until 6:00 am on Tuesday morning when the snow was 12 inches deep. Kevin
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Answer:

what is expected at 7am is 15 inches deep snow but what we have is 12 inches deep snow. The equation has failed in its prediction.

Step-by-step explanation:

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Read 2 more answers
Match each expression in the left column with the correct product in the right column.
lesya692 [45]

Answer:

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-6i

6i

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Step-by-step explanation:

1) √4 . √-3 . √-3

$ \sqrt{4} = 2 $

$ \sqrt{-3} . \sqrt{-3} = (\sqrt{-3})^2 $

$ \sqrt{4} . (\sqrt{-3})^2 = 2 \times -3 = $ -6

2) √-4 . √-3 . √-3

$ \sqrt{-4} = 2i $ .

Therefore, $ \sqrt{-4} . \sqrt{-3} . \sqrt{-3} = 2. \sqrt{-1}  \times -3 = 2i \times (-3) =  $ - 6i

3) √4 . √3 . √-3

$ \sqrt{4} = 2 $

$ \sqrt{3} . \sqrt{-3} = (\sqrt{3})^2 . \sqrt{-1} $

$ \implies 2 \times 3i = $ 6i

4) √4 . √3 . √3

$ \sqrt{4} = 2 $

$ \sqrt{3} . \sqrt{3} = (\sqrt{3})^2  = 3 $

Therefore, √4 . √3 . √3 = 2 . 3 = 6

5 0
3 years ago
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