Our aim is to calculate the Radius so that to use the formula related to the area of a segment of a circle, that is: Aire of segment = Ф.R²/2
Let o be the center of the circle, AB the chord of 8 in subtending the arc f120°
Let OH be the altitude of triangle AOB. We know that a chord perpendicular to a radius bisects the chord in the middle. Hence AH = HB = 4 in
The triangle HOB is a semi equilateral triangle, so OH (facing 30°)=1/2 R. Now Pythagoras: OB² = OH² + 4²==> R² = (R/2)² + 16
R² = R²/4 +16. Solve for R ==> R =8/√3
OB² = OH² +
Answer:
Step-by-step explanation:
Here time (t) is the independent variable; power (p) is the dependent one.
The graph of p = 100 - 4t is a straight line with slope -4. 100 is the vertical intercept. Battery power begins at 100 and steadily decreases after that.
Answer: C
Step-by-step explanation:
The answer is 29.4 grams.
How to solve:
1. Convert $15 to quarters. You can do this by multiplying 15 by 4. You use 4 because 4 quarters equal to $1. This will equal to 60 quarters.
2. Multiply both 5.76 and 6.25 by 60 to find the weight of $15 in quarters for today and before 1965.
Today: 5.76 g x 60 = 345.6 grams
Before 1965: 6.25 g x 60 = 375 grams
3. Now that you have the weight of quarters for both time periods, subtract! 375 grams - 345.6 grams = 29.4 grams
