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.
Answer:
2.83
Step-by-step explanation:
d= Square root of (x2 - x1 )^2 +( y2- y1)^2
from the point given you
x1= -6
y1 = -17
x2 = -8
y2 = -19
by applying the formula
Square root of ( -8 - (-6))^2 + ( -19 - (-17))^2
Square root of (-8+6) + (-19+17)^2
Answer: To get 24
Step-by-step explanation
What you would need to do is first multiply 12 x 5 would come out as 60.
Then from 60 you subtract 17 and 12 to get 5. Divide by the 5 that you just got from 17 and 12 and to get 20. Using the 4 add it to the 20 to get 24.
Answer:
566.4
Step-by-step explanation:
do multipilication
C^2-3cd-6cd+18d^2
=c(c-3d) -6d(c-3d)
=(c-3d) (c-6d)
