Mathematics: Compiled from the Best Authors, and Intended to be the Text-book of the Course of Private Lectures on These Sciences in the University at Cambridge, Volum 2W. Hilliard, 1808 |
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Side 6
... mile . PROBLEMS , PROBLEM I. To find the solidity of a cube . RULE . . * Cube one of its sides for the content ; that is , multiply the side by itself , and that product by the side again . * DEMONSTRATION . Conceive the base of the ...
... mile . PROBLEMS , PROBLEM I. To find the solidity of a cube . RULE . . * Cube one of its sides for the content ; that is , multiply the side by itself , and that product by the side again . * DEMONSTRATION . Conceive the base of the ...
Side 27
... miles , or its circumference 25000 miles ? Ans . 198943750 square miles . 4. The axis of a sphere being 42 inches , what is the con- vex superficies of the segment , whose height is 9 inches ? Ans . 1187 5248 inches . 5. Required the ...
... miles , or its circumference 25000 miles ? Ans . 198943750 square miles . 4. The axis of a sphere being 42 inches , what is the con- vex superficies of the segment , whose height is 9 inches ? Ans . 1187 5248 inches . 5. Required the ...
Side 29
... miles . Ans . 263858149120 miles . PROBLEM XIII . To find the solidity of a spherical segment . RULE . * To 3 times the square of the radius of its base add the square of its height ; then multiply the sum by the height , and the ...
... miles . Ans . 263858149120 miles . PROBLEM XIII . To find the solidity of a spherical segment . RULE . * To 3 times the square of the radius of its base add the square of its height ; then multiply the sum by the height , and the ...
Side 64
... drawn into a wire of of an inch in diameter ; what will the length of the wire be , allowing no loss in the metal ? Ans . 97784 797 yards , or 55 miles 984 797 yards . 31. If a heavy sphere , whose diameter is 4 64 . MATHEMATICS .
... drawn into a wire of of an inch in diameter ; what will the length of the wire be , allowing no loss in the metal ? Ans . 97784 797 yards , or 55 miles 984 797 yards . 31. If a heavy sphere , whose diameter is 4 64 . MATHEMATICS .
Side 113
... mile , or 880 yards ; then each ship observes the angle , which the other and the fort subtend , and they are found to be 85 ° 15 ′ and 83 ° 45 ' . What is the distance between each ship and the fort ? ZA 83 ° 45 ' ZB 85 15 Sum 169 00 F ...
... mile , or 880 yards ; then each ship observes the angle , which the other and the fort subtend , and they are found to be 85 ° 15 ′ and 83 ° 45 ' . What is the distance between each ship and the fort ? ZA 83 ° 45 ' ZB 85 15 Sum 169 00 F ...
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Mathematics: Compiled from the Best Authors, and Intended to be the ..., Volum 1 Samuel Webber Uten tilgangsbegrensning - 1808 |
Vanlige uttrykk og setninger
abscisses altitude axis azimuth base Ca² cask centre complement cone conjugate cosine course curve DE³ declination departure describe dial diameter diff difference of latitude difference of longitude distance divide draw the parallel drawn ecliptic ellipse equal equinoctial EXAMPLES feet figure find the rest frustum height Hence horizon hour angle hour lines hyperbola hypotenuse inches intersection LATITUDE SAILING length measure Mercator's meridional difference middle latitude miles multiply NOTE oblique circle opposite ordinates parabola parallel of latitude parallel sailing perpendicular plane sailing pole prime vertical primitive Prob PROBLEM projection Prop proportional Q. E. D. COR quadrant radius rectangle Required the content rhumb right ascension right circle right line rule secant segment Side AC sine sphere spheric triangle spindle square star station Stereographic Projection stile sun's tance tang tangent THEOREM vertical
Populære avsnitt
Side 3 - A sphere is a solid bounded by a curved surface, every point of which is equally distant from a point within called the center.
Side 147 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 8 - Take the length of the keel within board (so much as she treads on the ground) and the breadth within board by the midship beam, from plank to plank, and half the breadth for the depth, then multiply the length by the breadth, and that product by the depth, and divide the whole by 94; the quotient will give the true contents of the tonnage.
Side 59 - ... small statue, the head of which is 97 feet from the summit of the higher, and 86 feet from the top of the lower column, the base of which measures just 16 feet to the centre of the figure's base. Required the distance between the tops of the two columns ? Ans.
Side 61 - A gentleman has a garden 100 feet long, and 80 feet broad ; and a gravel walk is to be made of an equal width half round it ; what must the breadth of the walk be to take up just half the ground? Ans. 25-968 feet.
Side 63 - If a heavy sphere, whose diameter is 4 inches, be let fall into a conical glass, full of water, whose diameter is 5, and altitude 6 inches ; it is required to determine how much water will run over ? AHS.
Side 62 - Ans. the upper part 13'867. the middle part 3 '605. the lower part 2-528. QUES J. 48. A gentleman has a bowling green, 300 feet long, and 200 feet broad, which he would raise 1 foot higher, by means of the earth to be dug out of a ditch that goes round it: to what depth must the ditch be dug, supposing its breadth to be every where 8 feet ? Ans.
Side 21 - ... 07958 in using the circumferences ; then taking one-third of the product, to multiply by the length, for the content. Ex. 1. To find the number of solid feet in a piece of timber, whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches ; also, the length or the perpendicular altitude 24 feet.
Side 187 - AC 2AC nearly ; that is, the difference between the true and apparent level is equal to the square of the distance between the places, divided by the diameter of the earth ; and consequently it is always proportional to the square of the distance.
Side 29 - ... -5236, for the content. RULE II. To 3 times the square of the radius of the segment's base, add the square of its height ; then multiply the sum by the height, and the product by -5236, for the content.