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:
The quadratic curve has the best correlation to the given data.
Step-by-step explanation:
Enter the data into a spreadsheet or graphing calculator and try the different regression options to see which gives the highest R-value. Here, the quadratic regression does that.
i'm 75% sure the answer is C hope this helps
I think it’s 2,000 but again I could be wrong I need like the options
Answer:
x = 12
Step-by-step explanation:
(whole secant) x (external part) = (whole secant) x (external part)
(6+x) *6 = (4+23) * 4
(6+x) *6 = 27*4
6(6+x) = 108
Divide each side by 6
6+x =18
Subtract 6 from each side
6+x-6 =18-6
x = 12
