| John Hill - Arithmetic - 1716 - 496 pages
...Rule. We will (hew you how to prove the I(ule of Three, and fo conclude this" Rule: ^ If 4 Numbers be proportional, the ProducT: of the two Means, is equal to the Product of the two Extreams. Hence to prove your Work, multiply the 4th Number found, by the firft Number.and if that... | |
| John Hill - Arithmetic - 1764 - 424 pages
...Golden R ut E» We will fliew you how to prove the Rule i and fo conclude this rule. If 4 numbers be proportional, the product of the two means is equal to the product of the two extremes. Hence, to prove your work, multiply the 4th number found by the firft number; and if that product be,... | |
| John Hill - Arithmetic - 1765 - 428 pages
...Rut*. "We will fhew you how to prove the Rule of Three, and Co conclude this rule. If 4 numbers be proportional, the product of the two means is equal to the product of the two extremes. Hence, to prove your work, multiply the 4th number found by the firft number; and if that produft be... | |
| John Ballard - Gaging - 1812 - 276 pages
...10 24 '5 8 •. 3852 250 1926 Proof of the Golden-Rule. If four Numbers be proportional, the Produft of the two Means, is equal to the Product of the two Extreams. Hence to prove the Work, multiply the 4th Number found, by the firft Number ; and if that... | |
| Zadock Thompson - Arithmetic - 1826 - 176 pages
...geometrical proportion ; that is, 14 and 12 will be the two meant, and 2 the first extreme. Now since the product of the two means is equal to the product of the extremes, it is plain that if the product of the means be divided by one extreme, the quotient will... | |
| Francis Walkingame - 1833 - 204 pages
...from the mean. Thus, in 3, 6, 12, 24, 48 ; 12 X 12 = 3 X 48 =6X24 = 144. In an even number of terms, the product of the two means is equal to the product of the extremes, or of any two equidistant terms. Thus, in 32, 1 6, 8,, 4, 2, 1 ; 8 X 4 = 32 X 1 = 16 X 2... | |
| George Willson - Arithmetic - 1836 - 202 pages
...ratio, from the very '] -iinition of proportion, is common to the two consequents. To say then, that the product of the two means is equal to the product of the two extremes, is merely asserting that the products of the same factors are equal. This equality of the productof... | |
| George Willson - Arithmetic - 1838 - 194 pages
...the two middle terms and one of the extremes given, to find the other extreme. We have learned, that the product of the two means is equal to the product of the two extremes. We may, therefore, substitute this product ; and, dividing by the given extreme, find the other. Thus,... | |
| Charles Davies - Arithmetic - 1838 - 292 pages
...yd. 25 : yd. £ s d. 5 :: 2 34 5 25)£10 16* 8d 20 25)216(8s 200 16 12 25)200(8<Z 200 PROOF. § 143. The product of the two means is equal to the product of the extremes (see § 141). Hence, if either of these equal products be divided by one of the mean terms... | |
| Charles DAVIES (LL.D.) - Arithmetic - 1843 - 348 pages
...302s. 3d £I5 .2s Ans. £15 2s 3d. 4cwt. 4 3qr. 26ZJ. £9 2s 20 182s 12 19 7 133 4 2184 PROOF. § 143. The product of the two means is equal to the product of the extremes (see § 141). Hence, if either of these equal products be divided by one of the mean terms... | |
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