Step-by-step explanation:
If a variables varies jointly, we can just divide it by the other variables in relation to it.
For example, since p variables jointly as q and square of r, then

where k is a constant
First, let find k. Substitute p= 200
q= 2, and r=3.



Now, since we know our constant, let find p.

Q is 5, and r is 2.



Answer: All Real Numbers Are Solutions
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°
<span>It is D: n(n+4). I solved it by a long, but easy method. First, I made up the value of N, the width, and made it 3. I then added 4 to 3, which is 7. So the length is 7 and the width is 3. The area of this made up rectangle is 21. The only answer choice that has 21 as the area is D.</span>
C is the answer. The height of a cylinder is the perpendicular distance between the bases.