Answer:
(r o g)(2) = 4
(q o r)(2) = 14
Step-by-step explanation:
Given
Solving (a): (r o q)(2)
In function:
(r o g)(x) = r(g(x))
So, first we calculate g(2)
Next, we calculate r(g(2))
Substitute 9 for g(2)in r(g(2))
r(q(2)) = r(9)
This gives:
{
Hence:
(r o g)(2) = 4
Solving (b): (q o r)(2)
So, first we calculate r(2)
Next, we calculate g(r(2))
Substitute 3 for r(2)in g(r(2))
g(r(2)) = g(3)
Hence:
(q o r)(2) = 14
The correct option is "d".
Given that L = 0.8T²
length of pendulum = 30ft
L= 0.8T²
30 = 0.8T²
T² = 30 / 0.8
T² = 37.5
T = √37.5 = 6.1 seconds
<span>So, 6.1 is the closest to the period in seconds for a pendulum that is 30 ft long.</span>
Answer:
$0.13 per oz
Step-by-step explanation:
since you have $2.86, you have to split it into 22 ounces. You divide the 2.86 and the 22oz. and get 0.13
For this case we have the following equation of motion:
h (t) = -16t ^ 2 + 90t
Equaling the equation to zero we have:
-16t ^ 2 + 90t = 0
We look for the roots of the polynomial:
(t) (- 16t + 90) = 0
t1 = 0 (initial position)
t2 = 90/16 = 5.625 s (time it reaches the ground again)
Answer:
it will return to the ground in:
t = 5.625 s
Min:3
Q1:7.5
Med:10
Q3:13
Max:16
Hope this helps. Have a nice day. Please mark brainliest!
