So,
720 pt./hr.
2 pt. = 1 qt.
720/2 = 360
360 qt./hr.
Now, since there are 60 mins. in 1 hr., multiply the number by 60.
360(60) = 21600
720 pt./hr. = 21,600 qt./min.
<u>Given</u>:
Given that FGH is a right triangle. The sine of angle F is 0.53.
We need to determine the cosine of angle H.
<u>Cosine of angle H:</u>
Given that the sine of angle F is 0.53
This can be written as,
![sin F=0.53](https://tex.z-dn.net/?f=sin%20F%3D0.53)
Applying the trigonometric ratio, we have;
----- (1)
Now, we shall determine the value of cosine of angle H.
Let us apply the trigonometric ratio
, we get;
----- (2)
Substituting the value from equation (1) in equation (2), we get;
![cos \ H=0.53](https://tex.z-dn.net/?f=cos%20%5C%20H%3D0.53)
Thus, the cosine of angle H is 0.53
You would have to divide by 3.14 to find the radius squared, and to find the diameter, just add the radius plus the radius.
Answer:
volume=0.000064![m^{3}](https://tex.z-dn.net/?f=m%5E%7B3%7D)
mass=0.5888kg
Step-by-step explanation: