Answer:
Three Times.
Step-by-step explanation:
First and foremost a binary search requires that the array be sorted in ascending or descending order.
After the initial step the search goes to the middle most value
i.e. ![[0, 2,8,14,14,15,18,18,19,19]](https://tex.z-dn.net/?f=%5B0%2C%202%2C8%2C14%2C14%2C15%2C18%2C18%2C19%2C19%5D)
will be the second 14, which is the first iteration.
Since 14 is less than 19 the search will eliminate the lower half of the including 14, and will iterate again through the array:
![[15,18,18,19,19]](https://tex.z-dn.net/?f=%5B15%2C18%2C18%2C19%2C19%5D)
The array will go to the middle most value which is 18, second iteration. Since 18 is less than 19, the program will eliminate the lower half and be left with two values i.e. ![[19,19]](https://tex.z-dn.net/?f=%5B19%2C19%5D)
.
The search will run once i.e. third iteration, and will return key value found.
Answer:
When you flip over the the x-axis, the y-coordinate becomes the additive inverse. I.e. ; if I flipped over the x-axis, a (x,4) would become (x,-4). When you flip over the y-axis the x becomes the additive inverse. (x,4) will become (-x,4)
Step-by-step explanation:
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Have a good day, and good luck! ⭐
Answer:
A=40, B=60, C=80
Step-by-step explanation:
180/7 to find what "1" would be in the ratio then multiply 20*2, 20*3, 20*4.
Answer:
- hemisphere volume: 262 m³
- cylinder volume: 942 m³
- composite figure volume: 1204 m³
Step-by-step explanation:
A. The formula for the volume of a hemisphere is ...
V = (2/3)πr³
For a radius of 5 m, the volume is ...
V = (2/3)π(5 m)³ = 250π/3 m³ ≈ 261.799 m³
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B. The formula for the volume of a cylinder is ...
V = πr²h
For a radius of 5 m and a height of 12 m, the volume is ...
V = π(5 m)²(12 m) = 300π m³ ≈ 942.478 m³
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C. Then the total volume is ...
V = hemisphere volume + cylinder volume
V = 261.799 m³ +942.478 m³ = 1204.277 m³
__
Rounded to the nearest integer, the volumes are ...
- hemisphere volume: 262 m³
- cylinder volume: 942 m³
- composite figure volume: 1204 m³
_____
As a rule, you only want to round the final answers. Here, the numbers are such that rounding the intermediate values still gives the correct final answer. That is not always the case.
