Answer: The total number of logs in the pile is 6.
Step-by-step explanation: Given that a stack of logs has 32 logs on the bottom layer. Each subsequent layer has 6 fewer logs than the previous layer and the top layer has two logs.
We are to find the total number of logs in the pile.
Let n represents the total number of logs in the pile.
Since each subsequent layer has 6 fewer logs then the previous layer, so the number of logs in each layer will become an ARITHMETIC sequence with
first term, a = 32 and common difference, d = -6.
We know that
the n-th term of an arithmetic sequence with first term a and common difference d is
Since there are n logs in the pile, so we get
Thus, the total number of logs in the pile is 6.
Answer: c
given volume is 125 cc
cube root of 125 is 5
aka 5×5×5=125
and 6×5×5=150
hence answer is c
Answer:
C) (square root 2)/2
Step-by-step explanation:
In quadrant III, the sine of an angle is positive. The value will be ...
sin(Θ) = √(1 -cos(x)^2) = √(1 -(-√2/2)^2) = √(2/4)
sin(Θ) = (√2)/2
Question is: how many 84s will fit in 5376? Let's think about some easy multiples:
84 * 100 = 8400, so it's too big
84 * 10 = 840, so it might work
84 | 5376 | 10
-840
84 | 4536 | 10
-840
84 | 3696 | 10
-840
84 | 2856 | 10
-840
84 | 2016 | 10
-840
84 | 1176 | 10
-840
84 | 336
We can't fit any more 840 in 336, so we check how many 84s are in 336 and what's the remainder:
84 | 336 | 4
- 336
So there's no remainder. Now we add all the partial quotients to get the final result:
10 + 10 + 10 + 10 + 10 + 10 + 4 = <u>64
</u>It's correct, I checked it with calculator. I just hope you'll be able to read something from that, it's quite difficult to do partial dividing with no pencil and paper :)
The answer is 4.23 the second meteorite wieghts 4.23 pounds
