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Helga [31]
3 years ago
15

Is it is it possible to have 43 pencils and divide the pencils into groups with the same number of pencils with none left over

Mathematics
2 answers:
anastassius [24]3 years ago
6 0

we have given 43 pencils

we need to divide it equally so that none can be without groups.

we need to find the factors of 43.

we know that 43 is a prime number. which cannot be divided.so it is not possible to divide the 43 pencils with none left over .

sladkih [1.3K]3 years ago
5 0

TO find out the answer, we need to check whether 43 have any factors or not

43 is a prime number that is it have only two factors 1 and 43 . So either 1 will take whole 43 pencils or 43 will take one pencil each .

So it is not possible to divide 43 pencils into groups with the same number of pencils and none left over .

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3 0
3 years ago
What is the percent increase from 126 points to 48 points
svetoff [14.1K]
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2 years ago
The length of a photograph is 4 in. greater than its width. The area of the photograph is
Aleks [24]

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8 0
3 years ago
Given the arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103,...
omeli [17]

We are given

arithmetic sequence: u(1)= 124, u(2)= 117, u(3)= 110, u(4)=103

so, first term is 124

u(1)= 124

now, we can find common difference

d=u(2)-u(1)

d=117-124

d=-7

now, we can find kth term

u(k)=u(1)+(k-1)d

now, we can plug values

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u(k) must be negative

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u(k)=131-7k

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3 0
3 years ago
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