* Craig's answer is not reasonable because to add fractions the denominators must be the same.
** Total distance = 5/8 + 1/2 = 5/8 + 4/8 = 9/8 miles
*** Using the line number to prove the answer:
The line number that represents the problem is in the attached figure.
while the distance between 0 and 1 divided to 8 sections
to represent (5/8) count 5 sections from zero ⇒⇒⇒ point (a)
and to represent (1/2) it is the midpoint between 0 and 1 which mean it is 4 sections but it will be counted from point (a) so, adding 4 sections to point (a) the result will be the point (b)
So, counting from 0 to point (b) will give us 9 sections
and while one section represents (1/8)
So the total distance will be 9 * (1/8) = 9/8 which is agree with the result obtained before
v=gt,
Initial value when time t=0, v=g*0=0.
t=0 s, v=0 m/s means that object’s initial velocity 0 m/s.
v=gt, and g is a constant g=9.8 m/s². We can write v=9.8*t.
Rate of change 9.8 means the acceleration due to gravity is 9.8 m/s².
Answer: C. The initial value is 0. That means that object’s initial velocity 0 m/s. The rate of change is 9.8. That means the acceleration due to gravity is 9.8 m/s2.
Mason compared the number of free throws made to the number of free throws missed. The probability would actually be 2/5 becahse 18+12 is 30, giving you your denominator, then you made 12. So, simplifying 12/30 gives you your probability of 2/5.
Hope this helps you!
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.