Answer:
4
Step-by-step explanation:
• Divide the larger number by the smaller number
• If remainder is 0 then the smaller number is the HCF
• If not repeat with the remainder and smaller number until a remainder of 0 is reached.
1820 ÷ 204 = 8 remainder 188
204 ÷ 188 = 1 remainder 16
188 ÷ 16 = 11 remainder 12
16 ÷ 12 = 1 remainder 4
12 ÷ 4 = 3 remainder 0 ← remainder of 0 reached
Thus 4 is the HCF of 1820 and 204
Answer:
r. should be less than equal to 11
Option b. domain: {−12,−8, 0, 1} range: {0, 8, 12} is the right answer
Step-by-step explanation:
When the relation is given in the form of a table the x-values are the input to the function, hence domain and the y-values are the output of the function hence range.
Given table is:
[x] -12, -8, 0, 1
[y] 0, 12, 0, 8
The domain will be: {-12,-8,0,1}
And Range = {0,8,12}
Hence,
Option b. domain: {−12,−8, 0, 1} range: {0, 8, 12} is the right answer
Keywords: Functions, domain, range
Learn more about domain and range at:
#LearnwithBrainly
Answer:
![(a)\ Area = 195\pi](https://tex.z-dn.net/?f=%28a%29%5C%20Area%20%3D%20195%5Cpi)
![(b)\ Volume = 700\pi](https://tex.z-dn.net/?f=%28b%29%5C%20Volume%20%3D%20700%5Cpi)
Step-by-step explanation:
Given
![h = 12](https://tex.z-dn.net/?f=h%20%3D%2012)
--- lower diameter
--- upper diameter
Solving (a): The curved surface area
This is calculated as:
![Area = \pi l(r_1 + r_2)](https://tex.z-dn.net/?f=Area%20%3D%20%5Cpi%20l%28r_1%20%2B%20r_2%29)
Where
--- lower radius
--- upper radius
And
---- l represents the slant height of the frustrum
![l = \sqrt{12^2 + (5 - 10)^2](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%7B12%5E2%20%2B%20%285%20-%2010%29%5E2)
![l = \sqrt{12^2 + (-5)^2](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%7B12%5E2%20%2B%20%28-5%29%5E2)
![l = \sqrt{144 + 25](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%7B144%20%2B%2025)
![l = \sqrt{169](https://tex.z-dn.net/?f=l%20%3D%20%5Csqrt%7B169)
![l = 13](https://tex.z-dn.net/?f=l%20%3D%2013)
So, we have:
![Area = \pi l(r_1 + r_2)](https://tex.z-dn.net/?f=Area%20%3D%20%5Cpi%20l%28r_1%20%2B%20r_2%29)
![Area = \pi * 13(5 + 10)](https://tex.z-dn.net/?f=Area%20%3D%20%5Cpi%20%2A%2013%285%20%2B%2010%29)
![Area = \pi * 13(15)](https://tex.z-dn.net/?f=Area%20%3D%20%5Cpi%20%2A%2013%2815%29)
![Area = 195\pi](https://tex.z-dn.net/?f=Area%20%3D%20195%5Cpi)
Solving (b): The volume
This is calculated as:
![Volume = \frac{1}{3}\pi * h * (r_1^2 + r_2^2 + (r_1 * r_2))](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%2A%20h%20%2A%20%28r_1%5E2%20%2B%20r_2%5E2%20%2B%20%28r_1%20%2A%20r_2%29%29)
This gives:
![Volume = \frac{1}{3}\pi * 12 * (5^2 + 10^2 + (5 * 10))](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%2A%2012%20%2A%20%285%5E2%20%2B%2010%5E2%20%2B%20%285%20%2A%2010%29%29)
![Volume = \frac{1}{3}\pi * 12 * (25 + 100 + (50))](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%2A%2012%20%2A%20%2825%20%2B%20100%20%2B%20%2850%29%29)
![Volume = \frac{1}{3}\pi * 12 * (175)](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cfrac%7B1%7D%7B3%7D%5Cpi%20%2A%2012%20%2A%20%28175%29)
![Volume = \pi *4 * 175](https://tex.z-dn.net/?f=Volume%20%3D%20%5Cpi%20%2A4%20%2A%20175)
![Volume = 700\pi](https://tex.z-dn.net/?f=Volume%20%3D%20700%5Cpi)