Zach runs a 100-meter race. 7 seconds after the race started zach is 35 meters from the starting line and reaches his max speed; he runs at this max speed for the rest of the race. zach notices that he is 75 meters from the starting line 12 seconds after the race started. what is zach's max speed? 7 incorrect meters per second suppose zach runs for an additional z seconds after reaching his max how far will zach travel during those additional z seconds? 5z incorrect meters what is zach's distance from the starting line 7 + z seconds after the race started? 35+z incorrect meters what is zach's distance from the starting line x seconds after the race started (provided x ≥ 7 )? 35+7x incorrect meters

MaxSpeed=8 m/s. Additional distance during z seconds after top speed is 8*z meters. Now the distance at 7+z seconds after the race starts is D=( 35+8*z) meters. For x ≥ 7 the time of the race, then  D=(35+(x-7)*8) meters is the distance traveled during that time.

Explanation:

First we are interested in calculating the top speed, that is constant. We know at 35 meters he reaches the top speed and this happens 7 seconds after the race started. Also he is at 75 meters from the starting line 12 seconds after the beginning, then we can use the definition of speed, in this case is constant, to get its value:  Speed=change in distance / change in time

Then Speed= (75-35)meters/(12-7)seconds = 40m/5s=8m/s. That is the topspeed=8 m/s  

Now from the 35 meters he runs at constant speed, then if we are told he runs z seconds and we are asked the distance he runs in that time, we know the distance at constant speed is D=V*t,

then D= 8m/s * z seconds=8z meters.

But here since the run is a 100 meters-race we have an upper bound for z, here he has left (100-35)=65 meters to run at 8m/s, then he can maximum run additional 8,125 second or 15,125 seconds in total for the race.

That is D=8*z meters, with z < 8,125 seconds, this is the additional distance he can run after reaching his top speed.

Now the distance at 7+z seconds after the race starts is D=( 35+8*z) meters, the first 35 m he runs until reaching his top speed and the rest of the distance at this speed.

for x the total time of the run at a given point, we are asked the distance he has traveled at that time, then we are also told x ≥ 7, then  D=(35+(x-7)*8) meters, we subtract the 7 initial seconds because they are already counted in the initial 35 meters.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form