Answer:
The average yearly growth o
Explanation:
Given that the tree grew from 18 ft 35 1/2 feet in 3 1/2 years.
The growth within these years is:
35 1/2 - 18
= 35.5 - 18
= 17.5
Now, this averages:
17.5/3.5 (3.5 is the number of years)
= 5
The average is 5 ft per year
Answer:
Aaron needs <u>2 more rolls</u> to complete the path.
Step-by-step explanation:
Given:
Total rolls Aaron has = 4
Part of path covered by using 
of a roll = 
So, in order to find the number of rolls required to cover the complete path is given using the unitary method.
Rolls used for of a path =
Therefore, rolls used to cover the whole path is given by dividing the rolls used for one-eighth of the path and the path covered. This gives,
Now, rolls required to complete the path is 6. But Aaron has only 4 rolls.
So, he will need 6 - 4 = 2 rolls more to complete the path.
Answer:
C
Step-by-step explanation:
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
