It should be 44 if it goes up to 1,10. If it is 1,5 it should end up as 44 As well. Sorry if I’m wrong. Please do not attack me.
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad radius=\stackrel{}{ r} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ (x-9)^2+y^2=4\implies (x-\stackrel{h}{9})^2+(y-\stackrel{k}{0})^2=2^2~\hfill \stackrel{center}{(9,0)} \\\\\\ (x-\stackrel{h}{3})^2+(y-\stackrel{k}{2})^2=4\implies (x-\stackrel{h}{3})^2+(y-\stackrel{k}{2})^2=2^2~\hfill \stackrel{center}{(3,2)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Cqquad%20center~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%20radius%3D%5Cstackrel%7B%7D%7B%20r%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%28x-9%29%5E2%2By%5E2%3D4%5Cimplies%20%28x-%5Cstackrel%7Bh%7D%7B9%7D%29%5E2%2B%28y-%5Cstackrel%7Bk%7D%7B0%7D%29%5E2%3D2%5E2~%5Chfill%20%5Cstackrel%7Bcenter%7D%7B%289%2C0%29%7D%20%5C%5C%5C%5C%5C%5C%20%28x-%5Cstackrel%7Bh%7D%7B3%7D%29%5E2%2B%28y-%5Cstackrel%7Bk%7D%7B2%7D%29%5E2%3D4%5Cimplies%20%28x-%5Cstackrel%7Bh%7D%7B3%7D%29%5E2%2B%28y-%5Cstackrel%7Bk%7D%7B2%7D%29%5E2%3D2%5E2~%5Chfill%20%5Cstackrel%7Bcenter%7D%7B%283%2C2%29%7D)
Check the picture below.
well, its radius didn't change, anyhow, you know what to check out.
9514 1404 393
Answer:
22
Step-by-step explanation:
The average of the 6 even integers is the odd integer halfway between the first three and the last three. That average value is 126/6 = 21, so the 4th integer is 21+1 = 22.
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All 6 are {16, 18, 20, 22, 24, 26}.
Answer:
It is not greater than 1/2
Step-by-step explanation:
11/12-8/9 = 1/36
Answer:
48
Step-by-step explanation:
24
+<u>24</u>
48
- First, line up the numbers and write down the subtracting symbol or addition symbol.
- Second, add the ones place
- Then, add the tens place
- If numbers go onto the hundreds place then add that but if not then your done.
