The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
the answer to this question is 12 and 60
have a nice day
Answer:
Step-by-step explanation:
Good evening ,
I guess it’s the C expression,
Because, C. 3n^2+n-2 so the coefficient of the term n is 1 and the coefficient of the term n^2 is 3
:)
Answer:
x = 12
Step-by-step explanation:
sqrt(x+4) - 3 = 1
First get the sqrt all by itself on one side of the equation. Add 3 to both sides of the equation.
sqrt(x+4) = 1 + 3
sqrt(x+4) = 4
To "fix" the sqrt, that is, "undo it" and get rid of it, you have to SQUARE both sides of the equation.
(sqrt(x+4))^2 = 4^2
x + 4 = 16
subtract 4 to finish up.
x = 12
Check:
sqrt(12 + 4) - 3 = 1
sqrt16 - 3 = 1
4 - 3 = 1
1 = 1 Check!
Answer:
12
Step-by-step explanation:
3/4 is slope and 12 is why intercept
