Genet multiplied a 3-digit number by 1002 and got ab007c, where a, b, and c stand for digits. what was genet's original 3-digit number? genet multiplied a 3-digit number by 1002 and got ab007c, where a, b, and c stand for digits. what was genet's original 3-digit number?

Solution

Here we have to multiply  abc by 1002.

Then we have to positions the digits. We can able to position the digits by comparing the other digits.

In ones place, we have c

In the tens we have 7. in ab007c

By comparing the digits, we get

c = 9

b = 3

a = 5

Therefore, the 3-digit number is 539.

The answer is 539.

Here the resultant a, b, c stands for any digits.

to have more of what you want like money

step-by-step explanation:

save it by not spending on unnecessary stuff and have more

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15