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:
1) slope 3/2, y-intercept 7/2
2y-3x=7
2) slope 1/5, y-intercept 3
5y-x=15
3) slope 1/5, y-intercept 4/5
-x+5y=4
4) slope 5/2, y-intercept 7/2
2y=5x+7
Step by step explanation:
1) 2y-3x=7
2y=3x+7
y=3/2y+7/2
2)5y-x=15
5y=x+15
y=1/5x+3
3)-x+5y=4
5y=x+4
y=1/5x+4/5
4)2y=5x+7
y=5/2x+7/2
8.-12y^10 9.-15z^17 10.a^3b^3/125 11. 9x^2y^6/z^4
I think the answer would be c
Students in 2014 = (100% + 33%) × 2,300
students in 2014 = 133% × 2,300
students in 2014 = 133/100 × 2,300
students in 2014 = 3,059
There are 3,059 students in 2014