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>
For an inscribed angle like angle AHB, the measure will be half of the intercepted arc which we can see is 90 degrees.. therefore the measure of angle AHB is 45 degrees.
It would be log3 14/log3 4
Hope this helps. (:
Answer:
P(4≤x≤7) = 2/3
Step-by-step explanation:
We'll begin by obtaining the sample space (S) i.e possible outcome of rolling both dice at the same time. This is illustrated below:
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
Adding the outcome together, the sample space (S) becomes:
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
Next, we shall obtain the event of 4≤x≤7. This is illustrated below:
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
4 5 6 7
Finally, we shall determine P(4≤x≤7). This can be obtained as follow:
Element in the sample space, n(S) = 36
Element in 4≤x≤7, n(4≤x≤7) = 24
Probability of 4≤x≤7, P(4≤x≤7) = ?
P(4≤x≤7) = n(4≤x≤7) / nS
P(4≤x≤7) = 24/36
P(4≤x≤7) = 2/3
