1/2 = cos(2*t)sqrt(3) /2 = sin(-2*t)find tfind the time derivative of x(t), y(t)(-2sin(2*t) , -2cos(2*t) )plug in the t you found to find the velocity vector.
Answer:
<c = 45°
<d = 90°
I am not sure okay....this ia sq right?
<h2>
Answer:</h2>
The graph is shown in the Figure below
<h2>
Step-by-step explanation:</h2>
In this exercise, we have an equation. On the left side we have a straight line with slope 
and there is no any y-intercept. On the right side, on the other had, we also have a straight line, but the slope here is 
. Therefore, by plotting these two straight lines, we have that the solution is the origin, that is, the point 
.
Answer:
20 games won
Step-by-step explanation:
duh
Answer:
The equation of this line would be 4x + y = 13
Step-by-step explanation:
In order to find this equation we must first find the slope of the original line. To do this, we solve the original equation for y.
4x + y - 2 = 0
4x + y = 2
y = -4x + 2
The original slope (the coefficient of x) is -4, which means the new slope will also be -4 because parallel lines have the same slope. Now, we can use this slope along with the point in point-slope form to find the equation of the line. Just plug in the numbers and solve for the coefficient.
y - y1 = m(x - x1)
y + 3 = -4(x - 4)
y + 3 = -4x + 16
4x + y + 3 = 16
4x + y = 13
