If a number is a multipule of 6 then it is a multiplule of 2 AND 3
if a number is a multiplule of 2 then the last digit should be divisible by 2 (ie, 0,2,4,6,8)
if a number is divisible by 3, then the sum of its digits is divisble by 3
try each
A. 333
last number is not divisible by 2 so not divisible by 6
B. 882
2 is divisible by 2
8+8+2=18
18/3=6
divisible by 6 yes
C. 424
4 is disibielbe by 2
4+2+4=10
10/3=not a whole number so not divisible by 6
D. 106
6 is divisible by 2
1+0+6=7
7/3= not a whole number so not divibislb eby 6
answer is B
(10/17) / (-15/17) =
10/17 * - 17/15 =
- 10/15 =
- 2/5 <==
============
2.75 / -2.2 =
- 1.25 <==
============
(-2 3/5) / (3/5) =
- 13/5 * 5/3 =
-13/3 or - 4 1/3 <==
============
(2 1/4) / (3/4) =
9/4 * 4/3 =
9/3 =
3 <==
Answer:
this is the answer: 28
hope i helped :)
Step-by-step explanation:
Answer:
The correct option is C
Step-by-step explanation:
if f(x)= x3 + x2 - 20x
Replace f(x) by y
y = x3 + x2 - 20x
0 =x3 + x2 - 20x
x3 + x2 - 20x = 0
Take out x as a common:
x(x2+x-20)=0
Find factors of x2+x+20.
x(x^2+4x-5x-20) = 0
x{x(x+4)-5(x+4)}=0
x(x+4)(x-5)=0
Set x= 0
x=0 , x+4=0 , x-5 =0
x=0, x=0-4 , x=0+5
x=0, x= -4, x=5
x=(0,5,-4)
The correct option is C....
