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:
It's closer to 1
Step-by-step explanation:
Get the x terms by themselves on one side and a constant on the other side of the equal sign...
4x^2-8x=1 make the leading coefficient equal to one...
x^2-2x=1/4 now halve the linear coefficient, -2 in this case, square it, and add that value to both sides of the equation...-2/2=-1, -1^2=1 so
x^2-2x+1=1+1/4
x^2-2x+1=5/4 now the left side is a perfect square...
(x-1)^2=5/4 take the square root of both sides...
x-1=±√(5/4) add 1 to both sides
x=1±√(5/4)
Total surface area = 2 * circular base + circumference * height
= 2 * pi*(4)^2 + 8pi * 12
= 32pi + 96pi
= 128pi answer
Answer:
C and D. You're welcome :)
Step-by-step explanation:
