Answer:
64°
Step-by-step explanation:
∠RQP = 46°
∠PRQ = 180° - 110° = 70°
∠PRQ + ∠RQP + <u>∠RPQ</u> = 180°
70° + 46 + <u>∠RPQ</u> = 180°
116° + <u>∠RPQ</u> = 180°
Find <u>∠RPQ</u>
<u>∠RPQ</u> = 180° - 116° = 64°
To solve this we are going to use the formula for the volume of a sphere:

where

is the radius of the sphere
Remember that the radius of a sphere is half its diameter; since the first radius of our sphere is 24 cm,

. Lets replace that in our formula:



Now, the second diameter of our sphere is 36, so its radius will be:

. Lets replace that value in our formula one more time:



To find the volume of the additional helium, we are going to subtract the volumes:
Volume of helium=

We can conclude that the volume of additional helium in the balloon is
approximately <span>
17,194 cm³.</span>
19.......................................
The 4^10 and 4^5 can be simplified. So the answer is:
10^4 x 4^5
(The top right corner)