For this problem, you know that the first walker will arrive 2 hours before the second, and increases his speed by 2 times the second walker. You also know there is a distance of 24 km. So up until some time x, the two walkers have to be going the same speed. If the first walker increases speed by two times the speed per hour, and arrives two hours earlier, then his initial speed will be 20 km/h, because after 2 hours, he will have an increase of 4 km/hr, and the second will have an increase of 2 km/h, thereby making the first arrive 2 hours earlier, if that makes sense.
2 5/8 divided by 3
(2*8)+5/8 divided by 3
16+5/8 divided by 3
(3*7/2^3)(1/3)
3*7/2^3*3
7/2^3
7/8
each friend would get 7/8
Q is located at (-5,2)
Q" is located at (6,2)
they bot h have the same Y value (2) so we know it was only a reflection across the X axis
we need to find out where on the X axis it was reflected.
the distance between -5 and 6 = 11 units
11/2 = 5.5
the reflection line was 5.5 units to the right of the original Q
so the reflection line would be X = 0.5
Answer:
y=60x+14
Step-by-step explanation:
60 is the slope that is going UP at a certain rate and then ur adding 14
Answer:
B. 3.34s
Step-by-step explanation:
Find t when the object hit the ground, h(t)=0.
h(t)=-16t²+178
0=-16t²+178
16t²=178
t²=11.125
t=√11.125=3.33541601603s
