Using derivatives, it is found that:
i) 
ii) 9 m/s.
iii) 
iv) 6 m/s².
v) 1 second.
<h3>What is the role of derivatives in the relation between acceleration, velocity and position?</h3>
- The velocity is the derivative of the position.
- The acceleration is the derivative of the velocity.
In this problem, the position is:
item i:
Velocity is the <u>derivative of the position</u>, hence:
Item ii:
The speed is of 9 m/s.
Item iii:
Derivative of the velocity, hence:
Item iv:
The acceleration is of 6 m/s².
Item v:
t for which a(t) = 0, hence:
Hence 1 second.
You can learn more about derivatives at brainly.com/question/14800626
Answer:
X= 18
Step-by-step explanation:
18-18=0
x=18
By applying 2×pie×radius (radius+height)
This answer comes 1925000mm^3
Answer:
-6
-6i
6i
6
Step-by-step explanation:
1) √4 . √-3 . √-3
-6
2) √-4 . √-3 . √-3
.
Therefore, 
- 6i
3) √4 . √3 . √-3
6i
4) √4 . √3 . √3
Therefore, √4 . √3 . √3 = 2 . 3 = 6
Answer:
1.3
Friend is wrong
Step-by-step explanation:
Given:
friend's claim: height of his building is more than 1.50 times the height of yours
line of sight to the top edge of the other building makes an angle of 21° above the horizontal
line of sight to the base of the other building makes an angle of 52° below the horizontal
Solution:
Let A be the height of your building is A
Let B+A his building is B higher than yours.
Let the distance between the buildings is x.
then
tan 52 = A/x
tan 21 = B/x
A/B = tan 52 / tan 21
= 1.27994 / 0.38386
A/B = 3.33
(A + B) / A = 1.5
0
A/A + B/A = 1.50
1 + B/A = 1.50
B/A is basically (B/x) / (A/x)
So
1+ 3.33 / 3.33
= 4.33/3.33
= 1.3
Since 1.3 is not equal to 1.5
Hence the friend's claim is wrong.
