The answers are in the picture
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Answer:
2 unit/time²
Step-by-step explanation:
Given the equation:
v(t) =t^2-3t
At interval ; 1, 4
V(1) = 1^2 - 3(1)
V(1) = 1 - 3
V(1) = - 2
At t = 4
V(4) = 4^2 - 3(4)
V(4) = 16 - 12
V(4) = 4
Average acceleration : (final - Initial Velocity) / change in time
Average acceleration = (4 - (-2)) ÷ (4 - 1)
Average acceleration = (4 + 2) / 3
Average acceleration = 6 /3
Average acceleration = 2
Answer:
They are taking 12 2 credit courses
The are taking 4 1 credit courses
Step-by-step explanation:
x = 1 credit courses
y = 2 credit courses
The number of courses is 16
x+y = 16
The number of credits is 28 so multiply the course by the number of credits
1x+2y=28
Subtract the first equation from the second equation
x+2y =28
-x-y=-16
-----------------
y = 12
They are taking 12 2 credit courses
We still need to find the 1 credit courses
x+y = 16
x+12= 16
Subtract 12 from each side
x-12-12 = 16-12
x =4
The are taking 4 1 credit courses
If you would like to solve the system of equations, you can do this using the following steps:
-3x + 4y = 12
x * 1/4 - 1/3 * y = 1 ... x * 1/4 = 1 + 1/3 * y ... x = 4 + 4/3 * y
_____________
<span>-3x + 4y = 12
</span>-3 * (4 + 4/3 * y) + 4y = 12
-12 - 4y + 4y = 12
-12 = 12
-12 - 12 = 0
-24 = 0
The correct result would be: <span>the system of the equations has no solution; the two lines are parallel.</span>
