Answer:
The frequency of the note a perfect fifth below C4 is;
B- 174.42 Hz
Step-by-step explanation:
Here we note that to get the "perfect fifth" of a musical note we have to play a not that is either 1.5 above or 1.5 below the note to which we reference. Therefore to get the frequency of the note a perfect fifth below C4 which is about 261.63 Hz, we have
1.5 × Frequency of note Y = Frequency of C4
1.5 × Y = 261.63
Therefore, Y = 261.63/1.5 = 174.42 Hz.
Answer:
Option C is correct...hope this helps:)
Answer:
Part A: See the image attached. The angle between the radius and the chord will always be 90 degree if the radius is perpendicular to the chord
Part B: There are many ways to prove this and here is mine
Prove OC is perpendicular to AB
We know that ΔAEC≅AED by SSS criteria (SSS criteria is when all three sides of a triangle is equal making them congruent)
to prove this that they apply to the SSS criteria:
m∠ADO=m∠BDO
m∠ADO=90°=m∠BDO
Part C: See the image attached. A radius is perpendicular to a chord when the chord is equally divided in half by the radius.
Part D: There are many ways to prove this and here is mine
Prove OC is perpendicular to AB
AO=BO because AO and BO are both radii
OD≅OD by reflexive theorem
ΔADO≅ΔBDO because they are both right triangle with the hypotenuse and a side
AB≅BC means that OC is perpendicular to AB
If you are still confused then I recommend watching these vdos by khan academy about proving perpendicular radius bisector: https://youtu.be/2NgzruZ4_5c
https://youtu.be/TgDk06Qayxw
Hope this helped bye:)