What is the range of the relation?
1 answer:
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Answer:
From first Newton's equation of motion:
Substitute the variables:
Acceleration is 2 ms-²
Answer:
18 square units
Step-by-step explanation:
When one side is vertical or horizontal, it is convenient to find the height from the opposite vertex to that side. It is generally an integer number of grid squares.
Here, the height is 4 -(-2) = 6. The base length is 1 -(-5) = 6. That means the area is ...
A = 1/2bh
A = 1/2(6)(6) = 18 . . . . square units
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<em>Alternate solution</em>
If you want to go at this the hard way, you can find the length of LK and the distance of J to that line.
The length of LK is ...
d = √((x2 -x1)^2 +(y2 -y1)^2) = √((4 -(-2))^2 +(3 -(-5))^2) = √(36 +64) = 10
The distance to line LK can be found by writing the equation of the line LK in general form:
4x -3y -7 = 0
Then the distance formula for point (x, y) is ...
d = |4x -3y -7|/√(4^2 +3^2)
d = |4(-2) -3(1) -7|/5 = |-8 -3 -7|/5 = 3.6
Using the area formula with these dimensions gives ...
A = 1/2bh = 1/2(10)(3.6) = 18 . . . . square units
Answer:
x = 8
Step-by-step explanation:
Lines plotted through the given points intersect at (x, y) = (8, -4).
The x-coordinate of the solution is 8.
