Answer:
0
Step-by-step explanation:
(y2-y1)/(x2-x1)
6(y2)-6(y1)= 0
2(x2)- -3(x1)=5
0/5=0
<span>Use the graph to state the solution for the system.<span><span>x + 2y = 4 (line a)
</span><span> 3x – 2y = 4 (line b)</span></span><span><span /></span>C.
</span>
(2, 1)
Answer: The last one 15
+15g
Step-by-step explanation:
The last one , is not equivalent because after simplifying the expression, you will get 15g^2 + 45g and the last one doesn't match to it.
Answer:
a) clockwise sense
b) x² + y² = 1
c) y ∈ [0,1]
Step-by-step explanation:
a) x = cos (t) y = -sin (t)
for t ∈ [ 0 , 2π ]
for t = 0 point P is x = cos (0) = 1 y = -sin (0) y = 0
The point is on the x axis
Now let´s see what happens for t = π/4
x = cos (π/4) x = √2/2 and y = - sin (π/4) y = - √2/2
Note π/4 is 45°
The coordinates of the new point P ´(t = π/4) are in the fourth quadrant
that means the point is moving clockwise sense.
b) x = cos(t)
y = -sin(t)
Squaring on both sides of each equation we get:
x² = cos²t
y² = sin²(t)
Adding side by side
x² + y² = cos²t + sin²t
x² + y² = 1
Values for y are y ( 0 , 1)
Answer:
<h3>444m</h3>
Step-by-step explanation:
Given the height of the rocket modelled by the equation;
y=-16x^2+152x+83
The rocket reaches its maximum height when the velocity of the rocket is zero.
Since velocity is the change in displacement with respect to time;
velocity v = dy/dx
dy/dx = -32x + 152
at dy/dx = 0
-32x + 152 = 0
-32x = -152
x = -152/-32
x = 4.75s
To get maximum height reached by the ball, we will substitute x = 4.75 into the modeled function as shown;
y(4.75) = -16(4.75)^2+152(4.75)+83
y(4.75) = -361+722+83
y(4.75) = 444m
Hence the maximum height reached by the ball is 444m
