Answer:
Claim 2
Step-by-step explanation:
The Inscribed Angle Theorem* tells you ...
... ∠RPQ = 1/2·∠ROQ
The multiplication property of equality tells you that multiplying both sides of this equation by 2 does not change the equality relationship.
... 2·∠RPQ = ∠ROQ
The symmetric property of equality says you can rearrange this to ...
... ∠ROQ = 2·∠RPQ . . . . the measure of ∠ROQ is twice the measure of ∠RPQ
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* You can prove the Inscribed Angle Theorem by drawing diameter POX and considering the relationship of angles XOQ and OPQ. The same consideration should be applied to angles XOR and OPR. In each case, you find the former is twice the latter, so the sum of angles XOR and XOQ will be twice the sum of angles OPR and OPQ. That is, angle ROQ is twice angle RPQ.
You can get to the required relationship by considering the sum of angles in a triangle and the sum of linear angles. As a shortcut, you can use the fact that an external angle is the sum of opposite internal angles of a triangle. Of course, triangles OPQ and OPR are both isosceles.
The graph of a function h(x) translated n units up is represented by the following function:
Therefore, the graph of g(x) is the graph of f(x) translated 3 units up.
The graph of f(x) is:
Therefore, the graph of g(x) is:
Answer:
The volume πr^2*height/3=1471
π*r^2=1471/9*3
π*r^2=490.33(I'm rounding all my numbers to the hundredths digit)
r^2=490.33/3.14(taking 3.14 for pi)=about 156.16
r= sqrt 156.16=about 12.5
Again this answer is not precise but I rounded everything to the hundredths digit because pi is never ending. I hope it helped!
A cylinder has a cross section of a circle as well as the sphere.
