What is the area of a parallelogram whose vertices are A(−1, 12) , B(13, 12) , C(2, −5) , and D(−12, −5) ?
1 answer:
You might be interested in
something noteworthy is that the squared variable is the "x", thus the parabola is a vertical one, the "p" value is negative, so is opening downwards, and the h,k is pretty much the origin,
vertex is at (0,0)
the focus point is "p" or 5 units down from there, namely at (0, -5)
the directrix is "p" units on the opposite direction, up, namely at y = 5
the focal width, well, |4p| is pretty much the focal width, in this case, is simply yeap, you guessed it, 20.
Answer:
1. x > 4
2. m < 2
3. x > 2
4. x < -6
Step-by-step explanation:
These 4 inequalities will be solved the same way we solve equations. We take variables to one side and numbers to another and use algebra to solve. Each of them are solved shown below:
1.
so x is greater than 4, we can write (variable first):
x > 4
2.
so m is less than 2, we can write:
m < 2
3.
so x is greater than 2, we can write:
x > 2
4.
so x is less than -6, we can write:
x < -6