 | Charles Davies - Arithmetic - 1850 - 412 pages
...square AE, the two rectangles FE and EC, and the square ED. Hence, The square of two figures is equal to the square of the tens, plus twice the product of the tens by the units, plus the square of the units. Let it now be required to extract the square root of 1296.... | |
 | Rufus Putnam - Arithmetic - 1850 - 274 pages
...tens from the given cube, the remainder contains three times the square of the tens, plus three times the tens multiplied by the units, plus the square of the units, all multiplied by the units. (Why?) (97.) Therefore, dividing the cd 1744 remainder by the former factor,... | |
 | James Robinson (Jr.) - Arithmetic - 1851 - 336 pages
...period of figures to this remainder, and we hare 784 for a new dividend. This dividend contains twica the product of the tens multiplied by the units, plus the square of the units. Twice 4, (tens) the root already obtained, is 8, (tens) which is the greater factor of twice the product... | |
 | Daniel Leach - Arithmetic - 1851 - 280 pages
...(40)2~4800. 302. The complete divisor is always three times the square of the tens, phis three times the tens, multiplied by the units, plus the square of the units / as? 3 (40)2+3 (40x5)+52zz:5425. 303. The difference between any trial divisor and the next complete... | |
 | Charles Davies - 1852 - 344 pages
...sum is 32+2(3 x 6)+62 : that is, 3 + 6 3 + 6 32+3x6 3'+2(3x6)+6 H The square of a number is equal to the square of the tens, plus twice the product of the tens by the units, plus the square of the units. The same may be shown by the figure : Let the line AB re-... | |
 | Daniel Leach - Arithmetic - 1853 - 622 pages
...(40+5 )a= 3 (40)3 +6(40X5)+3(5)2=6075. It is evident, from the preceding propositions, that three times the tens multiplied by the units, plus the square of the units, added to any trial divisor, will give the next complete divisor. It is also evident, that the difference... | |
 | David Henry Cruttenden - Arithmetic - 1853 - 330 pages
...square of 45. Thus, 45 = 40+ 5 40+ 5 200 + 25 1600 + 200 Hence, 452 = 1600+400 +25=2025 Ans. That is, the square of the tens, plus twice the product of the tens into the units, plus the square of the units. 16. What is the 4th power of 12 ? Ans. 20,736. 17. Find... | |
 | William Smyth - Algebra - 1855 - 370 pages
...the tens by a and the units by b, we have a -j- b = 47, and squaring both sides a* -j- 2 ab -\-!>' = 2209. Thus the square of a number, consisting of units...2209, the square of 47, we have The square of the tens (a2) = 1600 Twice the tens by the units (2 a V) = 560 The square of the units (V) = 49 2209 Considering,... | |
 | Elias Loomis - Algebra - 1855 - 356 pages
...841 " 800+40 + 1. If, then, 841 is the square of a number composed of tens and units, it must contain the square of the tens, plus twice the product of the tens by the units, plus the square of the units. But these three terms are blended together in 841, and... | |
 | Arithmetic - 1856 - 46 pages
...itself is called the square, or second power, of that number. The square of any number is equal to the square of the tens, plus twice the product of...multiplied by the units, plus the square of the units. 1. What is the product of 125 multiplied by 125 ? Solution. 120 x 120 = 14400 + 120 x 5 x 2 = 15600... | |
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Which proves that the square of a number composed of tens and units contains, the square of the tens plus twice the product of the tens by the units, plus the square of the units. 
