For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.
Using the speed - distance relationship, the time left before the appointed time is 27 minutes.
<u>Recall</u><u> </u><u>:</u>
<u>At</u><u> </u><u>10mph</u><u> </u><u>:</u>
- Distance = 10 × (t + 3) = 10t + 30 - - - (1)
<u>At</u><u> </u><u>12</u><u> </u><u>mph</u><u> </u><u>:</u>
- Distance = 12 × (t - 2) = 12t - 24 - - - - (2)
<em>Equate</em><em> </em><em>(</em><em>1</em><em>)</em><em> </em><em>and</em><em> </em><em>(</em><em>2</em><em>)</em><em> </em><em>:</em>
10t + 30 = 12t - 24
<em>Collect</em><em> </em><em>like</em><em> </em><em>terms</em><em> </em>
10t - 12t = - 24 - 30
-2t = - 54
<em>Divide</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>-</em><em> </em><em>2</em>
t = 54 / 2
t = 27
Hence, the time left before the appointed time is 27 minutes.
Learn more : brainly.com/question/25669152
Mike is driving 20 miles per hour
If he is driving 10 miles in 30 mins
(1/2 an hour) if you times both by two it tells you the MPH that Mike is driving.
For a square, the area is the square of the side length (length x width = area, but in a square length = width). That means the side length is the square root of the area.
In this case the side length is the square root of 121, which is 11. Answer is C.
