A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
Answer:
D.
Bob should use the mean to make his selling price look like it's the greatest.
I would have to say -7(1/8)x-4/3=20.
The first and the second one both result in
-23 5/7
The last one results in -24 8/21
but the third one is -22 3/8.
So your answer is either the third of the last one. I assumed the third because, it is lower than all others and the first two are closer to 24 than they are to 22. It’s your choice on how to judge it.
Answer:
im pretty sure this is the answer :) ym = (y1 + y2)/2
Step-by-step explanation:
