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sergejj [24]
3 years ago
10

1. 10^7 = 10,000,000​

Mathematics
1 answer:
enyata [817]3 years ago
8 0

Answer:

True

Step-by-step explanation:

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Step-by-step explanation:

When the problem asks you to find the zeros all they are asking for is the solution so to solve all you need to do is set each individual piece equal to zero.

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(a) How many integers in the range 1 through 120 are integer multiples of 2, 3, or 5? keyboard_arrow_down Solution (b) How many
Umnica [9.8K]

Answer:

(a) 88 integers

(b) 92 integers

Step-by-step explanation:

(a) integers whose last digits are divisible by 2 are multiples of two or numbers whose digits ends with zero. So for number 1-120 , all the even numbers which are sixty in number are are multiples of two.

For  3, numbers whose digits sum is divisible by three are multiples of three. 3,6,9,12,15,18,21,24,27,30 are multiples of three from numbers 1-30. we have four 30s in 120. which means numbers of integers will be 10*4 = 40integers. However out of these numbers , half are also integers of 2 which reduces the number added to 20integers.

For 5, numbers whose digits ends with 5 or 0 are multiples of 5. this gives us 24 integers for 1-120. but out of these 24integers, 16 are common integers of 2 and 3 which reduces the number added to 8integers.

Thus from 1-120 the intergers of 2,3 or 5 = 60+20+8 = 88integers.

(b) if we are considering from numbers 1-140;

for 2 we wil have 70 integers,

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For 7, numbers 7,14,21,28,35,42,49,56,63,70 are multiples of three from 1-70. This pattern is repeated from number 71-140. hence we have 20 integers in all. However 12 of the multiples are also multiples of either 2 or 5 which reduces the number to 8 integers.

Thus from 1-140, the integers of 2, 5, or 7 = 70+14+8 = 92integers

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