21 x2+10x+21x55x-100x + 525x+804x3-630x -55 -100 804 216 to find x. 525 -630 root 216(.79128784748 .7 7.49 19.943 10.7 28.49 -35.057 7 7.98 25.529 6.6696.-91.84665 242.182794 946.8607701 111.09195214 Two farmers sell two sorts of corn: A sells 5 bushels; B rereceives in all for his $36, Now, says B to A, if we add the number of my bushels to the number of your dollars, the sum will be 28. Says A to B, and if I add the square of my dollars to the square of your bushels, the sum will be 424. How many bushels did B sell, and how many dollars did A receive? See Index. Ans. B sold 18 bushels, and A received $10. A Given 3 +2+3x2+4x2+5x2+6x-654321, to find x. 4 5 6 654321(8.9569 979.929 12259.6344 336 1043 12599.929 96532.5705 45.81 1021.887 13217.7510 1088.81 13621.816 109750.3215 46.62 1064.574 792.9263 1135.43 14686.390 110543.2478 419058.498|4 47.43 1107.990 796.141 1182.86 15794.380 111339.388 5527.16239 377.850317 412818.1183|3 33.414535 5566.9694 29.395994 418388.0877 4.018541 673.4017 3.779571 .238970 673.989 209977 A man and his wife could drink a barrel of beer in 15 days; but after drinking together 6 days, the woman alone drank the remainder in 30 days. In what time could either alone drink the whole barrel? Since the man and his wife had been drinking together 6 days, and they could drink the whole in 15 days, it is evident they drank or of the whole; and the woman drank 1-2 or 3 in 30 days, or the whole in 50 days, or 3 times in 150 days. And by the question, they could both drink the whole 10 times in 150 days, since the man alone could drink the whole 7 times in 150 days, or once in 213. C. Given 16x-20x+5x=.5, or -1.25x+.3125x.03125, to find x, which gives the sine of 69, and .5 being the sine of 30°. .2|7|2|1|2|16|9 In an isosceles triangle, there is given the sum of the base and side of the inscribed square 18 poles, and the difference of their areas 84 square poles. To determine the triangle and square. Put a 84, h=18, and x for the side of the inscribed square. Then will b-x be the base of the triangle by the question. And 2x the difference between the base and side of the square. b Then, by similar triangles, b-2x: b-x::x: perpendicular of the triangle. Consequently, double area; from which taking 2x the double area of the, (b-x)2 we have b=2xx-2x2 = 2a, by the question. Hence 52-72x+66x-3024. And x-7.534. 2721280859 213715 -3047 190485 .272125039 23230 -3047 21770 .27212199/2 1460 -15 1361 82 = : (b2 — 2bx+x2)x — (b — x) — 2x2 = 2a(b-2x), or b'x-2bx+ x3-2bx2+4x3-2ab4ax, or 5x3-4bx2 + (b2+4a)x=2ab. Ans. x= 7.534, &c. Hence 5x3 72x2 + 660x 2x2 2x2 = 3024. = Let the side of the square be 2x, and the base of the triangle 2x+2y. Then, by similar triangles, we have y: 2x ::x: +2x=the perpendicular. Now, by the ques+2xy=84, and 4x+2y-18, or y=9-2x; this value y tion, ; hence 2x3 y y = of y substituted in the other equation, we have 5x3—36x2 + 165x 378. Hence x-3767; then 2x 7.534 the side of the square, and 2x+2y or 10.465373 the base of the triangle. Otherwise, Put a = 18, b= = 84, and x for the base of the triangle. Then will a―x be the side of the square, and (a— x)2 its area. Again, by similar triangles, 2x-a:x:: a−x : ax-2.2 2x a lar of the triangle; and therefore its area is the perpendicu -(a-x)-b. This reduced is 2-39·6x2+ 585-6x=2937.6; in which x is = 10.4654 the base of the trian ax-x2 gle. Consequently = 26.9009 its perpendicular, and a-z Given x3-39.6x2+585·6x=2937·6, or x3-7·2x2+33x =75.6 to, find one value of x, in each equation. 4th. Div. 21-297/4 17 1918 4th. Divisor. 85-3617 4262 3d Divisor. 85.8996 51215 36 51190 -8.215....— 411 25 17 8 |