Answer:
125 times bigger.
Step-by-step explanation:
A great way to visualize this problem is to think of it as a volume problem itself. Another good way to think about it is to plug in numbers in its place.
Way #1
5*5*5 (or 5^3)=125
Way #2
(2*2*2)=8
(2*5)*(2*5)*(2*5)=1000
1000/8=125
Answer:
y² + 11y + 18
Step-by-step explanation:
y² + 9y + 2y + 18
y² + 11y + 18
The answer is
y= - x-9/3 I believe.
1- Subtract 9 from both sides.
-3y = x - 9
2- Divide both sides by -3
y = - x-9/3
Answer:
Number of small boxes is 50 and number of big boxes is 60
Step-by-step explanation:
let the number of small box represent x and the number of big box represent y
Hence total number of boxes = x + y = 110 ......................Eqn1
Total weight of boxes = 30x + 55y = 4800----------------Eqn2
Solving both equations simultaneously
From Eqn1, x = 110-y, substitute it into equation 2
Hence, 4800 = 30(110-y) + 55y
1500 = 25y
y = 60
X = 110 - 60 = 50
Number of small boxes is 50 and number of big boxes is 60
Answer:
Step-by-step explanation:
So, we’re probably looking at a n9-m0 cusp, since adding 0 to a number gives a sum lower than adding 9 to a slightly smaller number, something that isn’t true elsewhere in the cycle.
We need a n9 where the sum n+9 is divisible by 3. 9 is divisible by 3 already, so n must be divisible by 3 as well. So, it’s 0, 3, 6, or 9.
Taking as the most likely 3 (because a 39 year old is more likely than a 9 year old, 69 year old, or 99 year old to have a child who doesn’t already know their father’s age) we’ll try 39, 40. 3+9=12, 4+0=4, 12/3=4, this is a possible solution.
Do the others work?
0+9=9, 1+0=1, 9/4=/=1
6+9=15, 7+0=7, 15/4=/=7
9+9=18, 1+0+0=1, 18/4=/=1
1+2+9=12, 1+3+0=4, 12/3=4
there are at least two possibles. the father was correct, the father cannot determine his age from what she has been told, she must guess. From what she knows of him, is it more likely that he’s 39, or 129, or even older?
