What is the midpoint of the line segment whose endpoints are (-8, 12) and (-13, -2)?
2 answers:
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Answer:
(-1, -1) and (2, 2)
Step-by-step explanation:
Substitute for y in the first equation and solve the quadratic.
x = x^2 -2
x^2 -x -2 = 0 . . . . subtract x
(x -2)(x +1) = 0 . . . .factor
Solutions are the values of x that make these factors be zero:
x = 2, x = -1
Since y=x, the solutions are ...
(x, y) = (-1, -1) and (2, 2)
Answer:
4
Step-by-step explanation:
= 3.74165738677
≈4
the number 7 before de "." being greater than 5 rounds up to the next number, in this case 4
We can draw a unit circle as:
The sine is the vertical leg (or vertical projection) of the triangle formed by the angle.
The cosine will be the horizontal leg (or horizontal projection) of this same triangle.
The hypotenuse of this triangle will be the radius if the unit circle (r = 1).
a) The sine will be increasing from θ = 0° until it reaches its maximum at 90°.
Then it will be decreasing for angles between 90° and 270°.
Between 270° and 0° it will be increasing again.
b) The cosine has a maximum value for an angle of 0°.
It will decrease until 180°.
Then it will be increasing from angles between 180° and 360°.
c) We can see in the unit circle that the sine is 0 when the angle is one the horizontal axis, like when θ = 0° or θ = 180°.
The cosine will be equal to 0 when the angle is vertical: θ = 90° and θ = 270°.
Answer:
a) Increasing: (0°, 90°) and (270°, 360°)
Decreasing: (90°, 270°)
b) Increasing: (180°, 360°)
Decreasing: (0°,180°).
c) Sin(θ) = 0 for θ = 0° and θ = 180°
Cos(θ) = 0 for θ = 90° and θ = 270°
