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:
If you are looking for the slope-intercept equation, it's y=−3x−1. Hope you have a great rest of the day and stay safe, be kind, and go beyond Plus Ultra!
9514 1404 393
Answer:
g(f(7/3)) = 4 4/9
Step-by-step explanation:
g(f(7/3)) = g((7/3)²) = g(49/9) = 49/9 -1 = 40/9
g(f(7/3)) = 4 4/9
Step-by-step explanation:
10. |-41|= 41
12. -|1.5|= -1.5
9514 1404 393
Answer:
x = 10·cos(θ) -4·cot(θ)
Step-by-step explanation:
Apparently, we are to assume that the horizontal lines are parallel to each other.
The relevant trig relations are ...
Sin = Opposite/Hypotenuse
Cos = Adjacent/Hypotenuse
If the junction point in the middle of AB is labeled X, then we have ...
sin(θ) = 4/BX ⇒ BX = 4/sin(θ)
cos(θ) = x/XA ⇒ XA = x/cos(θ)
Then ...
BX +XA = AB = 10
Substituting for BX and XA using the above relations, we get
4/sin(θ) +x/cos(θ) = 10
Solving for x gives ...
x = (10 -4/sin(θ))·cos(θ)
x = 10·cos(θ) -4·cot(θ) . . . . . simplify
_____
We used the identity ...
cot(θ) = cos(θ)/sin(θ)
