Introduction to Linear DrawingCummings, Hilliard,, 1825 - 64 σελίδες |
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Σελίδα 50
... answer is , 2 ten thousandths and 3 tenths of a ten thousandth . Again , divide 154,3 by 21,26 . 21,26 ) 15430 ( 7,25 14882 5480 4252 12280 10630 1650 remainder . The answer is 7 hundredths , and 25 hundredths of a hundred . It is ...
... answer is , 2 ten thousandths and 3 tenths of a ten thousandth . Again , divide 154,3 by 21,26 . 21,26 ) 15430 ( 7,25 14882 5480 4252 12280 10630 1650 remainder . The answer is 7 hundredths , and 25 hundredths of a hundred . It is ...
Σελίδα 56
... the whole length of the walls , 6336 inches . Multiply this by the height , and you have the answer in square inches . 6336 1145 1145 84,6 2023 2023 38016 25344 6336 50688 536025,6 square inches . 2. A man wishes to cover the walls of a 56.
... the whole length of the walls , 6336 inches . Multiply this by the height , and you have the answer in square inches . 6336 1145 1145 84,6 2023 2023 38016 25344 6336 50688 536025,6 square inches . 2. A man wishes to cover the walls of a 56.
Σελίδα 58
... answer , 6391 square yards of surface . The superficies , or surface of a polygon , or a pyra- mid , is found by taking separately the surfaces of the triangles of which they are composed . 2. An irregular court has a quadrilateral ...
... answer , 6391 square yards of surface . The superficies , or surface of a polygon , or a pyra- mid , is found by taking separately the surfaces of the triangles of which they are composed . 2. An irregular court has a quadrilateral ...
Σελίδα 59
... answer . PROBLEM IV . To find the surface of a trapezium . ( fig . 3. ) RULE . Take half the sum of the two parallel sides , and multiply by the height . Example 1. A roof in the form of a trapezium , has one of its parallel sides 44,7 ...
... answer . PROBLEM IV . To find the surface of a trapezium . ( fig . 3. ) RULE . Take half the sum of the two parallel sides , and multiply by the height . Example 1. A roof in the form of a trapezium , has one of its parallel sides 44,7 ...
Σελίδα 62
... answer . Ans . 13,869 square yards . 7. How many bricks would the above wall require ? Ans . 10110,5 . PROBLEM II . To gauge a cask . RULE . Take the superficies of the base , and twice that of the centre at the bung hole , ( Prob . 3 ...
... answer . Ans . 13,869 square yards . 7. How many bricks would the above wall require ? Ans . 10110,5 . PROBLEM II . To gauge a cask . RULE . Take the superficies of the base , and twice that of the centre at the bung hole , ( Prob . 3 ...
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Arithmetick axis called circumference circumscribe column comma cone cord CORINTHIAN ORDER cubick feet Cut a circle cylinder decimal Describe a circle diagonal divide DORICK ORDER dotted line draw a perpendicular Draw a right draw an arc Draw an oblique draw right lines ellipse ENTABLATURE equal sides Example fillets find the surface four sides given point graduated semicircle half diameter halves height hexagon horizontal lines hundredths inches long intercolumniation Ionick isoceles triangle length lengthened mark its centre measure Modules monitor Monitorial School Multiply the base obtuse ogee parallel sides parallelogram parallelopiped PEDESTAL plane parallel preceding figures Prob PROBLEM proportions pupil quarter quarter-round radius Raise a perpendicular rectangle regular polygon right angled triangle right line drawn RULE sandths scalene scalene triangle sided polygon sixth class smaller sides solid contents square inches square yards string subtract superficies tenths thousandths three equal torus trapezium triangular prism Tuscan upright