By substituting the given values for x calculate the y value for the following equation y=x+2
1 answer:
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The graph from which the position of the point <em>A</em> can determined following
the multiplication with a scalar is attached.
Responses:
- If <em>A</em> is in quadrant I and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant III</u>
- If A is in quadrant II and is multiplied by a positive scalar, <em>c</em>, then c·A is in <u>quadrant II</u>
- If <em>A</em> is in quadrant II and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant IV</u>
- If <em>A</em> is in quadrant III and is multiplied by a negative scalar, <em>c</em>, then c·A is in <u>quadrant I</u>
Background information;
The question relates to the coordinate system with the abscissa represent the real number and the ordinate representing the imaginary number.
Solution:
If A is in quadrant I; A = a + b·i
When multiplied by a negative scalar, <em>c</em>, we get;
c·A = c·a + c·b·i
Therefore;
c·a is negative
c·b is negative
- c·A = c·a + c·b·i is in the <u>quadrant III</u> (third quadrant)
If A is quadrant II, we have;
A = -a + b·i
When multiplied by a positive scalar <em>c</em>, we have;
c·A = c·(-a) + c·b·i = -c·a + c·b·i
-c·a is negative
c·b·i is positive
Therefore;
- c·A = -c·a + c·b·i is in <u>quadrant II</u>
Multiplying <em>A</em> by negative scalar if <em>A</em> is in quadrant II, we have;
c·A = -c·a + c·b·i
-c·a is positive
c·b·i is negative
Therefore;
c·A = -c·a + c·b·i is in <u>quadrant IV</u>
If A is in quadrant III, we have;
A = a + b·i
a is negative
b is negative
Multiplying <em>A</em> with a negative scalar <em>c</em> gives;
c·A = c·a + c·b·i
c·a is positive
c·b is positive
Therefore;
- c·A = c·a + c·b·i is in<u> quadrant I</u>
Learn more about real and imaginary numbers here;
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