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:
12,400
Step-by-step explanation:
Let the current employees be x
65% increase in number = 0.65x
If the total number of employees after the increase is 20,460, then;
x + 0.65x = 20,460
1.65x = 20,460
x = 20,460/1.65
x = 12,400
Hence the current number of employees is 12,400
Answer:
y = x/2 + 5
Step-by-step explanation:
y = x/2 + 5
⇒ y = (1/2)x + 5
which satisfies the linear equation y = mx + b
where m is the slope and b is the y-intercept
11, 500 • 11 = 5,500
Hope this helps! Let me know if you need any further assistance!
X= -3
4(-3) = -12
-12-5= -17
