Part A.
What we can do to solve this problem is to assume that
the acceleration of Bryan is constant so that the velocity function is linear.
The standard form of a linear function is in the form:
y = m x + b
or in this case:
v = m t + b
where v is velocity and t is time, b is the y –intercept of
the equation
The slope m can be calculated by:
m = (v2 – v1) / (t2 – t1)
m = (12 – 15) / (7 – 4)
m = -1
Since slope is negative therefore this means the cyclist
are constantly decelerating. The equation then becomes:
v = - t + b
Now finding for b by plugging in any data pair:
15 = - (4) + b
b = 19
So the complete equation is:
v = - t + 19
This means that the initial velocity of the cyclists at t
= 0 is 19 km / h.
Part B. What we can do to graph the equation is to
calculate for the values of v from t = 0 to 12, then plot all these values in
the Cartesian plane then connect the dots.
Answer:
Step-by-step explanation:
The only pair of corresponding angles in the choices is angles 2 and 6.
Answer:
Answer:
Step-by-step explanation:
Given a curve defined by the function 2x²+3y²−4xy=36
The total differential of this function with respect to a variable x makes the function an implicit function because it contains two variables.
Differentiating both sides of the equation with respect to x we have:
4x+6ydy/dx-(4xd(y)/dx+{d(4x)/dx(y))} = 0
4x + 6ydy/dx -(4xdy/dx +4y) = 0
4x + 6ydy/dx - 4xdy/dx -4y = 0
Collecting like terms
4x-4y+6ydy/dx - 4xdy/dx = 0
4x-4y+(6y-4x)dy/dx = 0
4x-4y = -(6y-4x)dy/dx
4y-4x = (6y-4x)dy/dx
dy/dx = (4y-4x)/6y-4x
dy/dx = 2(2y-2x)/2(3y-2x)
dy/dx = 2y-2x/3y-2x proved!
I would say no.
Let me know if you got it right
