The National Arithmetic on the Inductive System: Combining the Analytic and Synthetic Methods : Forming a Complete Course of Higher Arithmetic |
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Αποτελέσματα 1 - 5 από τα 45.
Σελίδα 34
... left 3 tens ; and , passing to hundreds , we add in the left - hand figure 1 ,
reserved from the 12 tens , which with the other figures 1 and 2 and 1 = 5
hundreds , which , taken from 6 hundreds , leaves 1 hundred ; and 138 is the
answer sought .
... left 3 tens ; and , passing to hundreds , we add in the left - hand figure 1 ,
reserved from the 12 tens , which with the other figures 1 and 2 and 1 = 5
hundreds , which , taken from 6 hundreds , leaves 1 hundred ; and 138 is the
answer sought .
Σελίδα 46
The i we write in the quotient , and obtain as the answer required 2341 . In this
illustration , to render the explanation the more concise , the naming of the
denominations of the figures has been omitted . When , as in the operation
preceding the ...
The i we write in the quotient , and obtain as the answer required 2341 . In this
illustration , to render the explanation the more concise , the naming of the
denominations of the figures has been omitted . When , as in the operation
preceding the ...
Σελίδα 52
The last remainder , 1 , being multiplied by the divisor , 100 , and 64 , the first
remainder , added , we obtain 164 for the true remainder ( Art . 77 ) ; and for the
answer required , 5760 years . Hence , when the divisor contains one or more ...
The last remainder , 1 , being multiplied by the divisor , 100 , and 64 , the first
remainder , added , we obtain 164 for the true remainder ( Art . 77 ) ; and for the
answer required , 5760 years . Hence , when the divisor contains one or more ...
Σελίδα 56
To this partial product we add the multiplicand , since , as it stands , it represents
the product of the multiplicand by the 1 ten of the multiplier ; and obtain 43764 ,
the answer required . In the second operation , we add in the multiplicand taken ...
To this partial product we add the multiplicand , since , as it stands , it represents
the product of the multiplicand by the 1 ten of the multiplier ; and obtain 43764 ,
the answer required . In the second operation , we add in the multiplicand taken ...
Σελίδα 58
... multiplicand , two places to the left , so that the multipli8 4 8 4 918 Ans . cand ,
as it stands over this partial product , will represent the product of the multiplicand
by the 1 unit of the multiplier ; and adding these we have the answer required .
... multiplicand , two places to the left , so that the multipli8 4 8 4 918 Ans . cand ,
as it stands over this partial product , will represent the product of the multiplicand
by the 1 unit of the multiplier ; and adding these we have the answer required .
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acres added amount answer Arithmetic balance barrels base bill bought bushels called cent Change ciphers column common compound containing cords cost cube cubic decimal denominator denoting diameter difference discount Divide dividend division divisor dollars equal EXAMPLES exchange expressed factors feet figures four fourth fraction gain gallons give given grains Hence hundred inches interest least common length less loss mean measure method miles mixed months multiplicand Multiply NOTE obtain OPERATION paid payment period pounds premium present prime principal proportion purchased quantity quotient rate per cent ratio received Reduce remainder result root RULE scale share shillings side simple sold square subtract TABLE taken tens third thousand tons true units weight whole number worth write written yards