| Alexander Ingram - 1830
...the value of a load, or of 50 cubic feet of timber at all prices, from 6d. to 2s. each foot. PROB. I. **To multiply numbers by the rule. Set 1 on B opposite to the multiplier on A** ; then oppos; to the multiplicand on B will be the product on A. 1. Multiply 16 by 6. . Ans. 96. 2.... | |
| 1845
...hundred : again, suppose 1 at the joint ten, 1 in the middle is 100, and 1 or 10 at the end is 1000, &c., **the intermediate divisions on which complete the whole...on B is 24 on A. The slide thus set, against 7 on** B is 28 on A. 32 .9 10 12 15 25 36 40 48 60 100, &c., &c. TO DIVIDE NUMBERS UPON THE RULE. PROPORTION,... | |
| 1847 - 178 sider
...hundred : again, suppose 1 at the joint ten, 1 in the middle is 100, and 1 or 10 at the end is 1000, &c-, **the intermediate divisions on which complete the whole...system of its notation- - , TO MULTIPLY NUMBERS BY THE** RULESet 1 on B opposite to the multiplier- on A ; and against the number to be multiplied on B is the... | |
| 1849 - 403 sider
...: again, suppose 1 at the joint ten, 1 in the middle is 100", and 1 or 10 at the end is 1000, &.C., **the intermediate divisions on which complete the whole...6 by 4. Set 1 on B to 4 on A ; and against 6 on B** a £4 on A. The slide thus set, against 7 on B is 28 on A. 8 9 10 12 15 25 32 36 " 40 " 48 " 60 " 100,... | |
| Alexander Ingram - 1851
...value of a load, or of 50 cubic feet of timber, at all prices, from 6d. to 2s. each foot. PROB. I. **To multiply numbers by the rule. Set 1 on B opposite to the multiplier on A** ; then opposite to the multiplicand on B will be the product on A. 1. Multiply 16 by 6 Ans. 96. 2 23... | |
| William Templeton (engineer.) - 1855
...hundred: again, suppose 1 at the joint ten, 1 in the middle is 100, and 1 or 10 at the end is 1000, &c., **the intermediate divisions on which complete the whole...against the number to be multiplied on B is the product** oi1 A. Multiply 6 by 4. Set 1 on B to 4 on A ; and against 6 on B is 24 on A. The slide thus set, against... | |
| charles haslett - 1856
...hundred. Again, suppose 1 at the joint ten, 1 in the middle is 100, and 1 or 10 at the end is 1000, <fcc., **the intermediate divisions on which complete the whole...multiplied on B is the product on A. Multiply 6 by 4. Set** l on B to 4 on A: and against 6 on B is 24 on A. The slide thus set, against 7 on B is 28 on A 38 9... | |
| Charles Haslett - 1855
...hundred. Again, suppose 1 at the joint ten, 1 in the middle is 100, and 1 or 10 at the end is 1000, <tc., **the intermediate divisions on which complete the whole...notation. To Multiply Numbers by the Rule. Set 1 on** В opposite to the multiplier on A ; and against the number to be multiplied on В is the product on... | |
| John Frost - 1855 - 403 sider
...hundred : again, suppose 1 at the joint ten, 1 in the middle is 100, and 1 or 10 at the end is 1000, &c., **the intermediate divisions on which complete the whole...A; and against 6 on B is 24. on A. The slide thus** get, against 7 on B is 28 on A. 8 " 32 " 9 " 36 " 10 « 40 « 12 " 48 " 15 |« 60 " 25 " 100,&c.&c.... | |
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