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You will note that the above numerals are written such that the order of values is either descending or the same:
1943 = 1000 + 500 + 100 + 100+ 100+ 100+ 10+ 10+ 10+ 10+ 1 + 1 + 1 =M+D+C+C+C+C+X+X+X+X+I+I+I = MDCCCCXXXXIII
Later, another principle was developed in order to reduce the number of letters involved in expressing large numbers such as 1943 as Roman numerals. When the letter for a smaller numeral is placed before the letter for a larger numeral, we subtract the two numerals. For example, IV means 5—1=4, which is preferable to IIII. The number 1943 can now be expressed more briefly as M(CM)(XL)III, with the numerals that are enclosed in brackets being subtracted. Now remove the brackets to give 1943 = MCMXLIII.
Example. Write MCM as a number in present-day numerals.
MCM = M + (M — C) C written before M means we subtract, since C is
smaller than M
= 500 + (500 - 100)
The first 20 numbers in Roman numerals are
1 =l 6 = VI 11 = XI 16 = XVI
2 = II 7 = VII 12 = XII 17 = XVII
3 = III 8 = VIII 13 = XIII 18 = XVIII
4 = IV 9 = IX 14 = XIV 19 = XIX
5 = V 10 =X 15 =XV 20 =XX
When we solve an equation we obtain solutions, and these solutions are sometimes called the roots of the equation. For example, x = 4 is a root of the linear equation 2x — 3 = 5, and x = 3 and x = —4 are the roots of the quadratic equation x2 + x — 12 = 0.
References: Linear Equation, Quadratic Equations.
A rotation is a geometrical transformation that frequently occurs in everyday life. A rotation turns a shape around. An example is a child’s carousel at a fair. The carousel spins about its center 0, which is fixed (figure a). The carousel can turn clockwise or counterclockwise through a full turn or through an angle which may be part of a full turn. Amanda decides to ride on the pig P.
If the carousel turns counterclockwise about the center O through an angle of 90°, the pig finishes up at the point P'. When the pig is in its starting position at P we say it is the “object.” When the pig is in its finishing position at P! we say it is the “image.” This is written as P -> P'; in words we say P maps onto P'. The whole transformation is described in the following way:
P' is the image of P after a rotation of 90° counterclockwise about the point 0. This point 0 is called the “center of rotation” and 90° is the angle of rotation. The angle of rotation is regarded as a positive angle when it is measured counterclockwise and a negative angle when it is measured clockwise.
If the pig starts at P again and this time the rotation is 180° about 0, the image of P is as shown in figure b as P!. A rotation of 180° can be regarded as clockwise or counterclockwise.
After a rotation about 0 through an angle of 270° clockwise the image of P is P\ as shown in figure c. It can be seen that a rotation of 270° clockwise is the same as a rotation of 90° counterclockwise.
Example 1. Draw the image of triangle ABC after a rotation of —90° about the origin. Write down the coordinates of the images of the points A, B, and C.
Solution. The rotation of the triangle ABC about the origin is in a clockwise direction, because the angle of rotation is negative, and is achieved by rotating about 0 each of
the points A, B, and C in turn (see figure d). To rotate the point B, join the points 0 and B with a straight line OB and draw a circle with center at 0 and radius OB. The points B and B lie on this circle.
With a set square, or protractor, measure a clockwise turn of 90° from OB and draw the line OB1. Repeat the process for each of the points A and C to obtain the image points A! and C'. Now join up the points A\ B \ and C’ to make the image triangle. As you can see, we rotate the points of the shape one at a time and then join them up to obtain the complete image. An alternative method is to make a tracing of the figure and turn the tracing paper clockwise through 90° and trace the image triangle.
In some problems the center of rotation may be a point that is part of the shape that is being rotated, as in the following example:
A compound pendulum is rotated counterclockwise through an angle of 30° about a point that is the midpoint of one of its sides. The object and image are drawn in figure e.
In the next example you are given the object and the image shapes and asked to find the center of rotation and the angle of rotation.
Example 2. The shaded triangle ABC in figure f is rotated about a certain point and A’B’C’ is the image. Find the center of the rotation and also the angle of rotation.
Solution. Join A to its image point A! (see figure g). Construct the perpendicular bisector of the line AA! and call it p. Join C to its image point C’. Construct the perpendicular bisector of the line CC! and call it q. The point O where the two lines p and q meet is the center of rotation.