over the fame Space, (at 13 Miles a Minute) 37508547 : o: 12 which is above 100000. Years. Put x Question the 2d. anfwered by R. P. A Number of Crowns b = Then by the Question { yx+b= y 3 ; 666 95. and 2 -c=sx; b By the first Step x = ·and x2= y Substitute in the fecond Step for x and xx their Value; then y+ 1 Уу Reduction and Tranfpofition we have this Equation yo + 4ya + 2by3 +cy2 — sby=bb. Hence y=9, and x = 7. Q. E. J. and x = Question the 3d. answered by Mr. B. Suppose the two Numbers to be x and y, then by the Question xy=(a perfect Square, which let be): Now for z'any Square may be affumed that is divifible by y', viz. put z 8, y=2. Then is 8. But 8 x 2 x 2 x 2, is a perfect Square as required. This is one of thofe kind of Queftions called Diophantus's, or indeterminate, being capable of an indefinite Number of true Anfwers. Question the 4th. anfwered by J. T. Put Draw the Lines as in the Scheme (Fig. 2d) and the three Sides of the Triangle being given, find the Perpendicular By, and the Segments of the Bafe. AC = 67 = 2a; Ay = 24. 6 = c; Cv = 42. 4 = d. By=56.9b; Aw= x; Cvy. Then will wo= x+y-za. Now because the Triangle ADw is by the Question to be equal to half the Triangle ABC, = 1⁄2 ab; And for the fame Reafon EK ab give thefe Proportions. First, As b: And the fimilar Triangles CBy, CEK, and A By, ADP ad y ab d: =CK; y2—ad and wP= 302 ас Hence alfo Kv و 22-ac Again the Triangles K Ev, DwP, wSv, are all fimilar, because the Angles at v and w, are common to all; and the Angles at S, K, and P, are right therefore : (a). And by multiplying Extreams and Means, we have x2 y2 — acy2 — adx2 +a2cd=a2b2. and y’— there remains the Area of the Triangle ASD= abx+2aby-4a2b 4x+4x-8 a Furthermore As A P: Py ; AD :D B. and as A D:DB:: ASD: DSB. Confequently :: ASD (abx+2aby — 4a3b, 4x+4y-8a bx3-5abx+2bxy-2aby+4a b. Likewife as 4x+49 8a wv ( x+y−2a ) : Cv (y) :: usw (az). aby 4x+4y-8a• v S v S C = 4 From EC ab fubtract v SC, and 2? there remains the Area of the Triangle CSE= aby+2abx-4a2b. 4x+4y- 8a But as CK: Ky CEE B. () and as CE: E BCSE, ESB. Ergo as C K ab 4 ab 4' viz. bx3— sabx+ 2bxy— 2aby+ 4a3b+by2 + 2bxy— Saby zabx+4a2b, divided by 4x + 49 202 which reduc'd and tranfpos'd is 8ax + 10a2 + y2 = =8ay 4xy. Subftitute for y2 and fore found. And x2 −8ax +10a2+ =8a-4x1a2b2 + adx2 — arcd. x2 y, their Values bea2b2 + adx 2 — a2cd, x2 ac. This reduc'd to one Denomination, and 8a -4x, brought under the Root, gives x^ -8ax3 +10a2 acx2+8a2cx-10a3c + a2b2+adx2 -a2cd, divided by x2 ac. =√16a2b2x2 +16adx4 16a2cdx 2 - 64a3b2x — 64a2dx3 +64a3cdx + 64a4b2 +64a3dx2 64a4cd, divided by x2 ac; For the known Co-efficients of x and its Powers put Letters, and x4-fx3 +8x2+ bx − k =√mx4 — nx3 +px2-qxtr Square the Numerator of the first Part of the Equation, and take away the Radical Sign of the 2d Part, and multiply it by 2-ac, and tranfpofe the Terms, will give this Equation of the 8th. Power. In Numbers-x+536.x770350.6 +2208588.** +1417310844-111015653533 +155776050139.x2 +7348869315871.x = 191821297287673. and by converging Series is found 63.2, hence y = 50.6 Now x and y being found, and confequently wv=x+y — 2a = 46.8=n, and the Area of the Triangle Su =ab= 476.5=2; put wS=x, vS=y; then y 4, x 2 =p and x2 + y2 = 222: But x=2p and x2 = 4pp 4224 yv. Ergo 4p2+2 = n2 ; y = √ 'n2 — √ π2+ — APP = 23.6, and x = 40.4; and the feveral Fences are, Question the 5th answered by J. F. Put (in Figure the 3d.) AB=50.=2b; The Area p q s nis=x+xd 3y, the Triangle q s w = x + x ×ay confequently the whole Area pqwfnp CD=30=20; By the Property of the Ellipfis we have - CC and b:cb:25 y2 and x = √ c2 b2 — 25 b 2 y2 CC Therefore 4yc2 bz-gb2 yz + 49 √ c2b2-25 baya CC must be a MAXIMUM. 32c2bzyy-576b2yzy сс This in FLUXIONS is 32c2b2yу—1600b2у3 y + - 2√16c 2b2 y2 —14462 94 21/16c2b2 y2 — 400 b2 4 divide by y and tranfpofe one of the Terms, and then involve both Sides, and by multiplying crofsways we fhall have an Equation of the 10th Power, which being divided by boy6 and tranfpofed, will give this following, quadratick Equation, viz. 14400 14400 34 — 2176 c2 y2 + 48 c4 = =}±"} y2= in Numbers 44 - 34. No2 y4 168. 75 and by com pleating the SQUARE and extracting the Roor, 120.25 2.461 and by this y 17 * - 14, 4, ≈ — 21.8, 3y = 7.38, sy = 12.30 and hence the four Sides of the Trapezoid are wy 24.6 ; st == 14.76; ws = yt = 36.53 and its Area = 712. 416 Square Feet, 2.E. I Queftion the 6th anfwered by I. T. In the first Place from the Area of the Circle given, √3848.46 -7854 we must find its Diameter which is = (as is demonstrated in moft Mathematical Treatifes) =70. Feet, (fee Figure the 4th) then in the Triangle AxC, we have given Ax = Ca= 35. and AC = 62. to find the perpendicular s = 16. 3 (for xs and xm are fuppofed to be perpendicular to AC, BD.) alfo in the Triangle BD, we have BxDx=35. and BD = 64. to find x = 14.2; but the Angle mas Complement of AD to 18076: 6 becaufe the Angles at s and m, are right, and all the Angles of the Quadrilateral Snmx, are equal to four right Angles; fo in the Triangle mxs are given the two Sides xm, xs, and the contained Angle ms, to find ms= 18.8, and the Angles msx, smx 46d: 58 and 57: 62,and fubftracting each of them from :we have in the Triangle m ns all the Angles and the Side ms to find n = 13.3 and 522 = 10.6: Laftly AS AC 31. and Bm BD = 32.0 the Triangles AxC, BD being fofceles, Ergo, As + STAN 41.6 and Bm •mn Bn 18.7 and now we have in the four Triangles made by the four Sides of the Trapezium, two Sides and a contained Angle in each, to find the faid four Sides, which are, viz. AB = 41.3, BC 30.8; CD45.; DA = 68.5; = Question the 7th answered by R. F. Here is given firft the Latitude failed from 54d: o, and the Ship's Departure 44. 45 Miles; and her Difference of Longitude 74. 17 Miles (vide Figure the 5th) Put Length of the Arch of the Latitude arriv'd in, then |