This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1915 Excerpt: ... 42 and i+rmf+2-(2) From (1), 15?/ = 15 x + X!/. (3) Multiplying each member of (2) by 2(x + )y + J), 84j/ + 42 = 8ix + 42 + 4xy + 2x + 2y + 1.(4) Canceling 42 = 42, subtracting (3) x 4, and reducing, 22t/ = 26x + l. Multiplying (3) by 22, 15 22 y = 330 x + x 22 y. Substituting (5) in (6), 15(26x + l) = 330x + x(26x + ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1915 Excerpt: ... 42 and i+rmf+2-(2) From (1), 15?/ = 15 x + X!/. (3) Multiplying each member of (2) by 2(x + )y + J), 84j/ + 42 = 8ix + 42 + 4xy + 2x + 2y + 1.(4) Canceling 42 = 42, subtracting (3) x 4, and reducing, 22t/ = 26x + l. Multiplying (3) by 22, 15 22 y = 330 x + x 22 y. Substituting (5) in (6), 15(26x + l) = 330x + x(26x + l). 26x2-59x-15 = 0. (2x-5)(13x + 3) =0..-. x = f or--ft. Rejecting the negative value and substituting for x in (5), y =3. Hence, A traveled 2 miles per hour and B 3 miles per hour. 21. From 237. Ex. 20, pd = WD. Substituting for d and D, the values given in the figure, and for W, 360,000, the weight on the wall column. Then p equals the weight on the interior column..-. 12p = 4 x 360,000. Solving, p = 120,000. But, the weight on the fulcrum equals the sum of the weights on the wall column and on the interior column. Hence, the weight on the fulcrum equals 360,000 pounds plus 120,000 pounds, or 480,000 pounds. 22. Let x = number of feet per second sound travels, and y = number of feet per second projectile travels. 3300 3360., Then' x + lT ' () 5600 /3360 2240N, . From (1), 10,080 y + 10,080 x = 13xy, (3) From (2), 10,080 y-10,080 x = bxy. (4) (3)+ (4), 18xi/=20,160!/..-. x=1120. (3)-(4), 8x-/= 20,160 x..-. y = 2520. Hence, sound travels 1120 feet per second, and the average velocity of the projectile is 2620 feet per second. 23. Substituting in v = gt, the given values, 2010 _ 32.lo' Solving, t = 62.5. Hence, if it takes 62.0 seconds for the bullet to rise, it will take twice that time for it to rise and return to earth, or 125 seconds. Substituting in s = gt1 the value for g and the value for t just found, s = 62,812.5. Hence, the bullet will rise 62,812.5 foot..-. i = J(-n Vm!-80). The roots will be imaginary if the discrim...
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