In circle A as shown below, segment DC is tangent to circle A at D. Also, DC = 4 and BC = 2
1 answer:
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You have to complete the square to get it into vertex form. Do this by setting the function equal to zero and at the same time moving the constant over to the other side of the equals sign so you have this: 
. Now we can complete the square on the polynomial by taking half the linear term, squaring it, and adding it to both sides. Our linear term is 4. Half of 4 is 2, and 2 squared is 4, so we add 4 to both sides: 
and simplify to get 
. In this process we have created a perfect square binomial on the left, which happens to be 
. Now move the -6 back over by addition to get 
. The vertex is found at (-2, 6), the third choice down.
Refer to the diagram shown below.
Given:
m∠A = 19°
c = 15
By definition,
sin A = a/c
Therefore
a = c*sin A = 15*sin(19°) = 4.8835
cos A = b/c
Therefore
b = c*cos A = 15*cos(19°) =14.1828
Answer:
The lengths are 4.88, 14.18, and 15.00 (nearest hundredth)
Given:
m∠A = 19°
c = 15
By definition,
sin A = a/c
Therefore
a = c*sin A = 15*sin(19°) = 4.8835
cos A = b/c
Therefore
b = c*cos A = 15*cos(19°) =14.1828
Answer:
The lengths are 4.88, 14.18, and 15.00 (nearest hundredth)