Since it's a multiple of 24, it has to be a multiple of the factors of 24.
Factors of 24:
2,3,4,6,8,12
You can use some of this knowledge to help create the number.
Since the # needs to be a multiple off 2, the last digit needs to be an 8
All numbers that are multiples of 3 have the property that all of their digits added together have to be a number that is evenly divisible by 3.
so your number will look like:
_ _ _ _ _ 8
so start trying combinations for the other 5 digits that give you a number that is a multiple of 3: 3,6,9,12,15, ect. If you can't find one, then it's impossible
Answer:
C
Step-by-step explanation:
0 = 8 is a false statement which means the system has no solutions
Answer:
1.(1,5) and (2,6) , 6-5/2-1=1/1 m=1 ,y=1x+b
5=1(1)+b 4=b
y=x+4
2.(1,1) and (3,-8) -8-1/3-1=-9/2 m=-9/2 ,y=-9/2x+b
1=-9/2(1)+b b=11/2
y=-9/x+11/2
3.(2.-3) and (5,-2) m=1/3 ,y=1/3x+b
-3=1/3(2)+b -3=2/3+b
-3-2/3=b
b=-11/3
y=1/3x-11/3
4.(2,5)and (4,3) m=-1 y=-1x+b
5=-1(2)+b 5=-2+b
5+2=b
b=7
y=-1x+7
6.(-3,-5) and (-1,-3) m=2/2=1 y=1x+b
-5=1(-3)+b -5=-3+b
-5+3=b
-2=b
y=1x-2
Step-by-step explanation:
Given expression is
![\sqrt[4]{\frac{16x^{11}y^8}{81x^7y^6}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E%7B11%7Dy%5E8%7D%7B81x%5E7y%5E6%7D%7D)
Radical is fourth root
first we simplify the terms inside the radical


So the expression becomes
![\sqrt[4]{\frac{16x^4y^2}{81}}](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%5Cfrac%7B16x%5E4y%5E2%7D%7B81%7D%7D)
Now we take fourth root
![\sqrt[4]{16} = 2](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B16%7D%20%3D%202)
![\sqrt[4]{81} = 3](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B81%7D%20%3D%203)
![\sqrt[4]{x^4} = x](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7Bx%5E4%7D%20%3D%20x)
We cannot simplify fourth root (y^2)
After simplification , expression becomes
![\frac{2x\sqrt[4]{y^2}}{3}](https://tex.z-dn.net/?f=%5Cfrac%7B2x%5Csqrt%5B4%5D%7By%5E2%7D%7D%7B3%7D)
Answer is option B
The answer is C because it is at the first of four intervals between 0 and 1