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Solved Problem 12.3 A $4,000, 90 day, 14 percent note receivable in settlement of an account, dated June 1, is discounted at 14 percent on July 1. Compute the proceeds of the note. Solution: Principal $4,000.00 Interest Income (90 days, 14%) 140.00 Maturity Value $4,140.00 96.60 Discount (60 days, 14% of maturity value) Proceeds $4,043.40 Solved Problem 12.4 What are the entries needed to record the infor mation in Solved Problem 12.3 (a) on June 1 (b) on July 1 Solution: (a) Notes Receivable Accounts Receivable (b) Cash Interest Income Notes Receivable 4,000.00 4,000.00 4,043.40 43.40 4,000.00



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(a) P(x, y) A Pr(4 (b) P(x, y) A Pr(x, 12 (c) P(x, y) A Pr( 20 (d) P(x, y) A Pr(x, 1

x, y) y) x, y) y)

A figure has reflectional symmetry if it looks the same after being flipped across an axis of symmetry that runs through its center. As illustrated in Fig. 18-7, a figure can have (a) one, (b) several, or (c) no axes of symmetry.

Fig. 18-7





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(a) Since P is on the bisector 4 j A, it is equidistant from AB and AC. Since Q is on the bisectors 4 j4 it is of of B, 4 equidistant from AB and BC. Since R is on the bisectors of j A and jB, it is equidistant from AB, BC, and 4 AC. R is the incenter of ^ ABC, that is, the center of its inscribed circle. (b) Since P is on the ' bisector of AB, it is equidistant from A and B. Since Q is on the ' bisector of AC, it is equidistant from A and C. Since R is on the ' bisectors of AB and AC, it is equidistant from A, B, and C. R is the circumcenter of ^ ABC, that is, the center of its circumscribed circle.

4.10 Applying principles 1, 3, 6, and 7 In each part of Fig. 4-34, find two points equidistant from the ends of a line segment, and find the perpendicular bisector determined by the two points.

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Solved Problem 12.5 Bill s Bank received a $20,000, 90 day, 14.5 per cent note from Sun Town Company on account, dated November 1. Sun Town paid the note on the due date. (a) Show the entries for this transac tion. (b) Show the adjusting entry assuming December 31 is the last day of the accounting period

Fig. 4-34

(a) B and D are equidistant from A and C; hence, BE is the ' bisector of AC. (b) A and D are equidistant from B and C; hence, AD is the ' bisector of BC. (c) B and D are equidistant from A and C; hence, BD is the ' bisector of AC. A and C are equidistant from B and D; hence, AC is the ' bisector of BD.

The angles of any triangle may be torn off, as in Fig. 4-35(a), and then fitted together as shown in (b). The three angles will form a straight angle. We can prove that the sum of the measures of the angles of a triangle equals 180 by drawing a line through 4 one vertex of the triangle parallel to the side opposite the vertex. In Fig. 4-35(c), MN is drawn through B parallel to AC. Note that the measure of the straight angle at B equals the sum of the measures of the angles of ^ ABC; that is, a b c 180 . Each pair of congruent angles is a pair of alternate interior angles of parallel lines.

Fig. 4-35

An exterior angle of a polygon is formed whenever one of its sides is extended through a vertex. If each of the sides of a polygon is extended, as shown in Fig. 4-36, an exterior angle will be formed at each vertex. Each of these exterior angles is the supplement of its adjacent interior angle.

For 8-bit clean transport, UTF-8 is not encoded, but the UTF-16 family must be encoded in base64. For binary clean transports [e.g., HTTP (RFC 2616)], no content transfer encoding is necessary. Otherwise, security and interoperability considerations, the published speci cation, and applications that use this media type are the same as for text/xml.

Fig. 4-36

Thus, in the case of pentagon ABCDE, there will be five exterior angles, one at each vertex. Note that each exterior angle is the supplement of an adjacent interior angle. For example, mja mja 180 .

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