Dan earns £8.50 per hour. how much will he learn for 8 hours work?
2 answers:
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Answer:
see below
Step-by-step explanation:
Every vertex moves twice as far from the center of dilation as it is in the pre-image.
Perhaps the easiest image point to find is the one at lower left. In the pre-image it is 2 units left of the center of dilation, so the image point will be 2×2 = 4 units left of the center of dilation. It will be located at (-6, -2).
Other points on the image can be found either by reference to the center of dilation, or by reference to known image points. For example, the next point clockwise is 1 left and 4 up in the pre-image, so will be 2 left and 8 up from (-6, -2) in the image. That is, the pre-image point (-5, 2) becomes image point (-8, 6). You will note that (-5, 2) is 3 left and 4 up from the center of dilation, and that (-8, 6) is 6 left and 8 up from the center of dilation (twice as far away).
Answer:
the comparison must be reversed when multiplying by a negative
Step-by-step explanation:
The rules of equality apply to solving inequalities, with the exception that multiplication or division by a negative number reverses the sense of the comparison:
-x > 1
x < -1 . . . . . multiply both sides by -1
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Effectively, multiplication (or division) by a negative number is equivalent to reflection across the origin. Things that were ordered left/right (on the number line) are ordered right/left after such a reflection:
-2 < -1
1 < 2
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<em>Additional comment</em>
Application of any function requires that you pay attention to ordering. Some functions naturally reverse the order; others do so only on specific domains.
Consider f(x) = 1/x.
1 < 2
f(1) > f(2) . . . . because the slope of the f(x) function is negative everywhere.
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<em>In the first and second quadrants</em>, the cosine function also reverses order.
20° > 10°
cos(20°) < cos(10°)
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Probably the most commonly encountered function used with inequalities is the absolute value function.
|x-3| > 2
This function has one domain where the slope is negative, and another domain where the slope is positive.
For x < 3, the function negates its argument, so we have ...
-(x -3) > 2
-x +3 > 2
-x > -1
x < 1 . . . . . everywhere consistent with x < 3
For x ≥ 3, the function does nothing, so we have ...
x -3 > 2
x > 5
The solution to this absolute value inequality is (x < 1) ∪ (x > 5).
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You can also resolve negative coefficients by adding the opposite. Addition and subtraction never require any change to the comparison operator.
-x > 1
0 > x +1 . . . . x is added
-1 > x . . . . . . -1 is added
