Hello :
<span>x² + y² - 12 x - 2 y + 12 = 0
(x²-12x) +(y² -2y) +12 = 0
</span>(x²-2(6)(x)+6²)-6² +(y² -2y+1) -1+12 = 0
(x-6)² +(y-1)² = 5²
<span>the center of a circle is (6, 1)</span>
You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.
So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:
C (m,n) = m! / (n! * (m -n)! )
=> C (50,30) = 50! / (30! (50 - 30)! ) = (50!) / [30! (50 - 30)!] = 50! / [30! 20!] =
= 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.
Answer:
y=1/2+1
Step-by-step explanation:
1) Y2-Y1 over X2-X1
<u>9-5=4</u>
12-4=8
4/8 simplify the answer= 1/2
m=1/2
2) subtitute the simplification to one of the points.
y=mx+b
5=1(4)+b
5=4=b
1=b
y=1/2+1
Answer:First one is parallel
Step-by-step explanation: