Answer: Krista can fill 3 1/3 bowls of gelatin.
Answer:
After 3 seconds while going up and 6 seconds while coming down
Step-by-step explanation:
Since projectile is launched from ground level with an initial velocity of v 0 feet per second, initial height =0
Height is given by
where v0 = initial velocity = 144
a) When h =288 ft.
At t= 3 or 6 seconds the projectile would be at height 288 ft.
So while going up after 3 seconds it wouldbe at a height of 288 ft and while coming after 6 seconds.
The function is divided by 12, you would find it out by dividing 120 by 10 and getting 12, if you wanted to check your answer, you would do 84 divided by 7 which is 12. the answer to 12 is 1 and 24 is 2. hope this helped :))
7/2= radius (3.5)
V=πr2^<span>h
V=</span><span>π3.52*12
V= 461.81 </span>
You have to make a cos graph that starts its minimum and of -2, has an amplitude of 10, a period of 10 and a maximum of 18.
I decided to use a cos graph since cos graphs start at their minimum or maximum unlike a sine graph that starts halfway between the minimum and maximum. You also know the amplitude has to be 10 since 18+2=20 and 20/2=10. We were also told that the water wheel completels a rotation every 10 minutes which means the period is 10 minutes.
lets start of with a regular cos(x) graph. This starts on its maximum instead of minimum so we have to multiply it by -1 to get -cos(x) which does start on its minimum.
-cos(x) has an amplitude of 1 instead of 10, to fix that we multiply it by 10 to get -10cos(x) which has an amplitude of 10.
-10cos(x) has a period of 2π instead of 10, to fix this we multiply the x by 2π/10 to get -10cos((π/5)x) which now has a period of 10.
-10cos((π/5)x) has a minimum of -10 and maximum of 10 instead of a minimum of -2 and maximum of 18, to fix this we add 8 to -10cos((π/5)x) to get -10cos((π/5)x)+8 which does have a minimum of -2 and maximum of 18.
Therefore the answer is y=-10cos((π/5)x)+8. x being time in minutes and and y being the height in feet.
I hope this helps. Let me know if anything is unclear.
