9514 1404 393
Answer:
1. ∠EDF = 104°
2. arc FG = 201°
3. ∠T = 60°
Step-by-step explanation:
There are a couple of angle relationships that are applicable to these problems.
- the angle where chords meet is half the sum of the measures of the intercepted arcs
- the angle where secants meet is half the difference of the measures of the intercepted arcs
The first of these applies to the first two problems.
1. ∠EDF = 1/2(arc EF + arc UG)
∠EDF = 1/2(147° +61°) = 1/2(208°)
∠EDF = 104°
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2. ∠FHG = 1/2(arc FG + arc ES)
128° = 1/2(arc FG +55°) . . . substitute given information
256° = arc FG +55° . . . . . . multiply by 2
201° = arc FG . . . . . . . . . subtract 55°
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3. For the purpose of this problem, a tangent is a special case of a secant in which both intersection points with the circle are the same point. The relation for secants still applies.
∠T = 1/2(arc FS -arc US)
∠T = 1/2(170° -50°) = 1/2(120°)
∠T = 60°
Answer:
The answer is the 1st one.
Step-by-step explanation:
Hope it helps! :)
Answer:
Sure, what do you need help on?
Answer:
We are given
Scientist begins with 150 milligrams of a radioactive substance
so,
we can use formula
we can plug Po=150
After 29 hours, 75 mg of the substance remains
so, at t=29 , P=75
we can plug it
now, we can solve for k
now, we can take ln on both sides
now, we can plug it back
and we get
we can plug t=33
we get
..............Answer
Step-by-step explanation:
