Answer:
Distance traveled by hockey puck of second player is 40 feet more than that of first player.
Step-by-step explanation:
Maximum height of 3 feet is attained by the hockey puck of first player is when the puck is 55 feet away from the player.
It is at the mid point of the total horizontal distance to be traveled by the hockey puck.
So, the horizontal distance traveled by hockey puck of first player = 55 + 55 = 110 feet
Second player:
![y=x(0.15-0.001x)](https://tex.z-dn.net/?f=y%3Dx%280.15-0.001x%29)
where
is the height of a second hockey puck
is the horizontal distance.
When the puck hits the ground, ![y=0](https://tex.z-dn.net/?f=y%3D0)
Now, let us find the value of
.
![\Rightarrow 0=x(0.15-0.001x)](https://tex.z-dn.net/?f=%5CRightarrow%200%3Dx%280.15-0.001x%29)
At
feet or
feet, ![y=0](https://tex.z-dn.net/?f=y%3D0)
Therefore, the distance traveled by second hockey puck is 150 feet.
So, the answer is:
<em>Distance traveled by hockey puck of second player is 40 feet more than that of first player.</em>
<em></em>
90 because there are 5 time as many candies with nuts as candies without nuts and there is 18 so it will be 18×5=90
Answer:
gotta be C
Step-by-step explanation:
The answer from least to greatest 5/8 0.62 0.615
Answer:
Well I think it is A because domain is the x values.
Step-by-step explanation:
So when you plug this in your calculator (mine is a ti-84 plus ce) you would hit graph. After it graphs it press Zoom, 0 to center it then press 2nd, trace which pulls up parabola menu's. Press 0 and find the left bound, right bound and then press enter which would give you x values of 2.9375 < t< 6
At the same time I don't know if this is right. I never really excelled at parabolas just trying to help.