Problem 1)
Minor arc DG is 110 degrees because we double the inscribed angle (DHG) to get 2*55 = 110
Answer: 110 degrees
=================================================
Problem 2)
Central angle GDC is the same measure as arc GFC. The central angle cuts off this arc.
The arcs GEC and GFC both combine to form a full circle. There are no gaps or overlapping portions.
So they must add to 360 degrees
(arc GEC) + (arc GFC) = 360
(230) + (arc GFC) = 360
(230) + (arc GFC)-230 = 360-230
arc GCF = 130
Answer: 130 degrees
=================================================
Problem 3)
Similar to problem 1, we have another inscribed angle. ABC is the inscribed angle that cuts off minor arc AC
So by the inscribed angle theorem
arc AC = 2*(inscribed angle ABC)
3x+9 = 2*(3x-1.5)
Solve for x
3x+9 = 2*(3x-1.5)
3x+9 = 6x-3
9+3 = 6x-3x
12 = 3x
3x = 12
3x/3 = 12/3
x = 4
If x = 4, then
arc AC = 3x+9
arc AC = 3*4+9
arc AC = 21
Answer: 21 degrees
=================================================
Problem 4)
Since we have congruent chords, this means that the subtended arcs are congruent. In this case, the arcs in question are CO and HZ
So arc CO is congruent to arc HZ
Answer is choice D
=================================================
Problem 5)
We have a right triangle due to Thale's theorem
The angles 75 degrees and x degrees are complementary angles. They must add to 90
x+75 = 90
x+75-75 = 90-75
x = 15
Answer: Choice D) 15
the answer is d
hope this helps
9514 1404 393
Answer:
D None
Step-by-step explanation:
The steepest line shown is line A, which has a rise/run = 5/2. It is not a slope of 4.
No line shown has a slope of 4.
This is just a line there are infinitely many solutions or points...for every value of x there is a value for y which is three times x. Pick any x solve for y and that is one of the infinite number of ordered pairs produced by the equation.
y=3x (0,0),(1,3),(2,6),(3,9).....................
