Answer:
<h2>A</h2>
Step-by-step explanation:
y > 0 - the region above the line together with that line
y ≥ 0 - the region above the line together with that line
y < 0 - the region below the line without that line
y ≤ 0 - the region below the line together with that line
We have:
y > 2 - the region above the line together with that line
y ≥ 2x - 3 - the region above the line together with that line
<em>look at the picture</em>
The common part is the solution.
They are the same slope
they are negative inversees (they multily to get -1)
2
-1/2
use the square viewer (on TI)
The relationship between the slopes of two lines that are parallel is they are the same.
The relationship between the slopes of two lines that are perpendicular is they are negative inverses of each other (they multiply to -1).
A line that is parallel to a line whose slope is 2 has slope 2.
A line that is perpendicular to a line whose slope is 2 has slope -1/2.
What must be done to make the graphs of two perpendicular lines appear
to intersect at right angles when they are graphed using a graphing
utility?
Answer:
E
Step-by-step explanation:
Answer:
3^6
Step-by-step explanation:
3^2 * 3^4
You add the exponents together.
Answer:
m∠H = 10°
Step-by-step explanation:
ΔFGH is an isosceles triangle. I made a diagram, so m∠F ≅ m∠H by B.A.T(base angle theorem). This is because an isosceles triangle have to congruent sides and angles.
Diagram is attached:
-Chetan K
