Ben and Josh went to the roof of their 40-foot tall high school to throw their math books off the edge.
1 answer:
Answer:
- Josh's book lands first
- Ben's lands about 0.648 seconds later
Step-by-step explanation:
Using the given equation for v=60 and s=40, the height of Ben's book is ...
h(t) = -16t² +60t +40
We want to find t when h(t) = 0, so we're looking for the solution to ....
0 = -16t² +60t +40
Using the quadratic formula, we find the positive value of t to be ...
t = (-60 -√(60² -4(-16)(40)))/(2(-16)) = (15 +√385)/8 ≈ 4.3277
__
Similarly, the height of Josh's book is ...
0 = -16t² +48t +40
t = (-48 -√(48² -4(-16)(40)))/(2(-16)) = (12 +√304)/8 ≈ 3.6794
__
The time before Josh's book lands is shorter by ...
4.3277 -3.6794 ≈ 0.6482 . . . . . seconds
Josh's book reaches the ground first, by about 0.648 seconds.
You might be interested in
Answer:
Step by step explanation :
Given:
it is given that in a rectangle the area is is 12x^2 -29x + 15
and the width is 4x- 3
Solution :
The rest is given in pic.
Thank u, please please mark it is Brainliest!
Hope u submit on time!
Answer:
1) 1 element
2) 13 elements
3) 22 elements
4) 40 elements
Step-by-step explanation:
1) Only one element will have no tails: the event that all the coins are heads.
2) 13 elements will have exactly one tile. Basically you have one element in each position that you can put a tail in.
3) There are elements that have exactly 2 tails. From those elements we have to remove the only element that starts and ends with a tail and in the middle it has heads only and the elements that starts and ends with a head and in the 11 remaining coins there are exactly 2 tails. For the last case, there are
possibilities, thus, the total amount of elements with one tile in the border and another one in the middle is 78-55-1 = 22
4) We can have:
- A pair at the start/end and another tail in the middle (this includes a triple at the start/end)
- One tail at the start/end and a pair in the middle (with heads next to the tail at the start/end)
For the first possibility there are 2 * 11 = 22 possibilities (first decide if the pair starts or ends and then select the remaining tail)
For the second possibility, we have 2*9 = 18 possibilities (first, select if there is a tail at the end or at the start, then put a head next to it and on the other extreme, for the remaining 10 coins, there are 9 possibilities to select 2 cosecutive ones to be tails).
This gives us a total of 18+22 = 40 possibilities.