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
1.5x + 0.2y = 2.68....multiply by 0.3
1.6x + 0.3y = 2.98...multiply by - 0.2
------------------------
0.45x + 0.06y = 0.804 (result of multiplying by 0.3)
- 0.32x - 0.06y = - 0.596 (result of multiplying by - 0.2)
----------------------add
0.13x = 0.208
x = 0.208/0.13
x = 1.6
1.5x + 0.2y = 2.68
1.5(1.6) + 0.2y = 2.68
2.4 + 0.2y = 2.68
0.2y = 2.68 - 2.4
0.2y = 0.28
y = 0.28/0.2
y = 1.4
solution (they intersect at) (1.6,1.4)
I believe the answer is a.
Answer:
B. x < -8
Step-by-step explanation:
Well first we need to get x by itself.
To do that we do 4 / -1/2 = -8
And since a negative was divided the > changes to a <.
So x<-8.
If x is less than -8 the line starts at -8 and goes to the left.
<em>Thus,</em>
<em>the answer is choice b.</em>
<em />
<em>Hope this helps :)</em>
Answer:
<h2><em><u>2a</u></em></h2>
Step-by-step explanation:
(a+b-c)-(b-a-c)
= a + b - c - b + a + c
= a + a + b - b - c + c
= <em><u>2a (Ans)</u></em>
