| Jeremiah Day - Algebra - 1820 - 352 pages
...terms, x : 1 4 : : 4 : 7 Dividing terms, x -. 2 : : 4 : 1 Therefore, x = 8. 6. If the number 20 l,e divided into two parts, which are to each other, in the duplicate ratio of 3 to 1, what number is a m.ean proportional between those parts ? Let x=. the greater part, and 20—... | |
| Bewick Bridge - Algebra - 1821 - 284 pages
...60ж + 900 = 900 — 800 = 1 00, and x— 30= + 10; or x= 30 ±10=40 or 20 the [parts required. Ex.2. The number 20 is divided into two parts, which are...Find a mean proportional between those parts. Let x= great er part, then 20 — x= lesser part; .'. by the question, x : 20 — x :: 3" : 1' :: 9 : 1. Hence,... | |
| Bewick Bridge - Algebra - 1821 - 648 pages
...THEOR. 12, a :d :a' : a •l\ Moreover, a : // but i/ : c by THEOR. 12, a : e :a3 : a :l,l :a4 Ex.2. The number 20 is divided into two parts, which are...other in the duplicate ratio of 3 : 1. Find a mean pro* portional between those parts. Let x= greater part, then 20— x = lesser part; .*. by the question,... | |
| Peter Nicholson - Mathematics - 1825 - 1046 pages
...a^— бОт+доО =900-800= 100, and x— 30= + 10; or лг=30+ 10=40 or '¿0 the parts required. Ok -The number 20 is divided into two parts, which are to each ofher in the duplicate ratio of 3 : 1 . It is required to find a mean proportional between those parts.... | |
| Bewick Bridge - Algebra - 1828 - 260 pages
...division^* — 60x= — 800 ; Л Xs— 60Х + 900 = 900-800=10О, andж— 30 = ±10 ; or x=30± 10=40 or 20 the [parts required. Ex. 2. The number 20 is...20 ~x= lesser part; .'. by the question, x: 20— ж:: 3* : l*::9: 1. Hence, by THEOH. 1, x= 180 — Qx, or 10ж=180; .'. x= 1 8 =± greater part. and... | |
| Jeremiah Day - Algebra - 1827 - 352 pages
...Adding terms, x : 1 4 : '. 4 : 7 Dividing terms, x : 2::4:1 Therefore, a-=8. 6. If the number 20 be divided into two parts, which are to each other in the duplicate ratio of 3 to 1, what number is a mean proportional between those parts ? Let*= the greater part, and 20 — x=... | |
| Bewick Bridge - Algebra - 1832 - 220 pages
...Sídivisíon^x" — 60ж= — 800 ; .•. Xя — 60ж+900=900 — 800=100, and x— 30= + 10, or ж=30+ 10=40 or 20, the parts required. Ex. 2. The number 20 is...between those parts. Let x— greater part, then 20 — x=lesser part ; .-. by the question, x : 20— ж :: 3a : Iя :: 9 : 1. Hence by THEOR. 1, x=180... | |
| Bewick Bridge - Algebra - 1839 - 280 pages
...division, x2 — 60x— — 800 ; .-. Xs— 60x+ 900= 900— 800=100, and x — 30=±10; or x=30±10=40 or 20, the parts required. Ex. 2. The number 20 is...Find a mean proportional between those parts. Let x=greater part, then 20 — x= lesser part ; .-. by the question, x : 20— x : : 32 : I2 : : 9 : 1.... | |
| Bewick Bridge - Algebra - 1841 - 260 pages
...division, x2 — 60x= — 800 ; .-. x2— 60x+900=900— 800=100, andx — 30=±10; or a:=30±l 0=40 or 20, the parts required. Ex. 2. The number 20 is...x= lesser part ; .-. by the question, x : 20— x : : 32 : I2 : : 9 : 1. Hence, by TH. 1, x =180 — 9x, or 10x=180; .-. x=18=greater part, and 20 —... | |
| James Bates Thomson - Algebra - 1844 - 272 pages
...shall be to the quotient of the less divided by the greater as 16 to 9. Prob. 6. If the number 20 be divided into two parts, which are to each other in the duplicate ratio of 3 to 1, what number is a mean proportional between those parts 1 Prob. 7. There are two numbers whose... | |
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