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5. If, in the figure at the left, the radius of the large circle is R and the radius of each of the smaller circles is r, show that the shaded area, A, is given by the formula
A=+(R+2 r)(R–2 r).
ameter is D and whose
inner diameter is d. If
the length is 1, show that the volume V occupied by the iron is expressed by the formula ·
_ D+ /D-.
Fig. 37. 7. The result in Ex. 6 may be expressed in words as follows: “To find the volume of a circular tube, multiply 7 times the length by half the sum of the two diameters by ——." Finish the sentence. Does this give us a convenient rule for determining any such volume ? Explain.
8. One side of a right triangle measures 12 feet, while the hypotenuse is 2 feet longer than the other side. Find the length of the other side.
V = all
For further exercises on this Chapter, see Appendix, pp. 298–301.
DIVISION AND FACTORING +
69. Division. The process of finding one factor when the product and other factor are given is called division.
The dividend is the given product, the divisor is the given factor, and the quotient is the required factor.
Thus, to divide 6 ab by 3 a means to find the number which when multiplied by 3 a gives 6 ab. Here the dividend is 6 ab, the divisor is 3 a, and the answer (or quotient) is evidently 2 b.
70. Law of Signs for Division. In § 27 we saw that the sign of a quotient is + whenever the dividend and divisor have like signs, and is – whenever they have unlike signs. This may be stated in a table as follows:
+ab :(+b)= ta
tab =(-6)=-a 71. Law of Exponents for Division. In dividing x5 by x3 we may proceed, as in arithmetic, by removing equal factors from dividend and divisor, the work appearing as below.
25 d. .* • * = 7(5-3) = x2. Ans.
23 . Similarly, we see that x6 = x2 is equal to xr6–2), or x4. In general, we have the following rule.
† 8 27 should be read again at this point,
RULE FOR FINDING THE QUOTIENT OF Two POWERS OF THE SAME NUMBER. The exponent of the quotient of two powers of the same number is equal to the exponent of the dividend diminished by that of the divisor. Stated in a formula, the rule becomes
Formula VIII. *"=x(m,n).
When m=n this formula gives x"/x" = x(n_n) = 20. But, we also have x"/x"=1 because numerator and denominator here are the same. Therefore, the meaning of xo must always be taken as 1; that is,
x0 =1. The pupil should carefully compare Formulas I and VIII.
ORAL EXERCISES State the quotient for each of the following indicated divisions. 1. 5) – 10 7. 6 ab. 11. a. 15. (p?9)". 2. –5)15
(1 p?q)3 3. – 3) – 15 8. -4 4. –2)10 6. 35, . Tabia
(a+b)3" 6. 2. 10. 14. (3.xy)? 18. (x-y+z).
*** (3 xy)3 (x-y+z)"
72. To Divide a Monomial by a Monomial. EXAMPLE 1. Divide 35 cAd2 by 5 cd. SOLUTION.
5 cd)35 c4d2
7 cd. Ans. Note that the process consists in first dividing the 35 by 5 to
obtain the new coefficient 7, then dividing the c4 by c, giving ? (by Formula VIII), then dividing the d2 by the d, giving d.
CHECK. (5 cd)X(7 c2d)=35 c4d2.
2 x2z)-10 xöy3z4
–5 xy3z3. Here the process is the same as in Example 1, but it is to be observed that the y3 which occurs in the dividend occurs without change in the quotient since the divisor contains no y factor.
CHECK. (2 x 2)X(-5 xy323)=-10 x3y3z4.
A careful examination of the two examples above enables one to work all similar examples without difficulty.
WRITTEN EXERCISES Find the quotient for each of the following indicated divisions, and check each answer.
1. –4 ab)28 a2b3 2. 3 xy2)9 x3y322
– 7 gk
4 xyz 27 mön p.
- 3 map 4 a4b3c5
20 a2bc3 [Hint. The coefficient of the quotient is no, or }.]
– 4 x?y425
14. (1) 3(*)2
[Hint. ==1. See $71.] 16. 52inly
73. To Divide a Polynomial by a Monomial. EXAMPLE 1. Divide 8 xy–4 x2y2 by 4 xy. Solution.
4 xy)8 x2y—4 xy2
2 x-xy. Ans. Note that the process consists in dividing each term of the polynomial, as in § 72, keeping due account of the connecting signs.
CHECK. 4 xy(2 x_xy)=8 x2y—4 x?y.
-3 a2bc2 – 4 a2 + 5 bcd. Ans. CHECK. -366-3 a2bc2–4 a +5 bed)=9 a2b2c2+12 a2b, 15 bcd.
WRITTEN EXERCISES Find the quotient in each of the following and check your answer. 1. 8 a462)24 a%b2+32 a5b3 – 16 a4b3 6 x?yz+12 xy22 — 24 xyz2.
-3 xyz mönp+mnöp.
i -a+a2b-a’c— add+ae