Let`s assume that points M, N and P are the touching points of those 3 circles:Then:Y M + M Z = 14,Z N + N X = 20X P + P Y = 18And also: M Z = ZN, Y M = P Y and N X = X P.Now we have a system of 3 equations ( Y M, M Z and X P are the radii of each circle ):Y M + M Z = 14M Z + X P = 20X P + Y M = 18 Y M - M Z = - 14+X P + Y M = 18 X P - M Z = 4Y M - M Z = - 14+M Z + X P = 20 X P - Y M = 6 /* ( - 1 )X P - M Z = 4 X P + Y M = - 6 X P - M Z = 4 Y M - M Z = - 2 Y M + M Z = 14 2 Y M = 12 => Y M = 6M Z - 6 = 2 => M Z = 8X P + 6 = 18
X P = 12
Radii of the circles are: 12, 8 and 6.
Answer:
2 4/25
Step-by-step explanation:
First, I made the decimal into 2 16/100. Then, keep simplifying the 16/100 until it is fully simplified.
The factors of 7are -1 and 7 or 1 and -7, the factors of 14 are 1, 2, 7, and 14, or -1, -2, -7,-14. so the list of potential zeros are: 1/1, 1/2, 1/7, 1/14, 7/1,7/2, 7/7, 7/14, which can be simplified into 1, 1/2,1/7, 1/14, 7, 7/2
add the negative ones: -1, -1/2,-1/7, -1/14, -7, -7/2
I believe there are a total of 12 potential zeros
reference:
http://www.sparknotes.com/math/algebra2/polynomials/section4.rhtml
