Algebra Demystified 2nd Ed
406 algebra De mystif ieD Let t represent the number of hours Gary needs to unload the truck by himself. John needs 36 minutes ...
Chapter 11 QuaDraTiC appliCaTionS 407 5 t − 12 = 0 3t + 1 = 0 (This does not lead to a solution.) 5 t = 12 t = 12 5 = 225 Gary n ...
408 algebra De mystif ieD when h = 0, so h 0 = −16t^2 + 1600 becomes 0 = −16t^2 + 1600. We solve this for t. 016 1600 16 1600 10 ...
Chapter 11 QuaDraTiC appliCaTionS 409 ✔SOLUTIONS For all of these problems, both a negative t and a positive t will be solutions ...
410 algebra De mystif ieD t^2 =^70 16 t^235 8 = ^ t =^35 8 t ≈ 2.09 The ball will hit the ground in about 2.09 seconds. For the ...
Chapter 11 QuaDraTiC appliCaTionS 411 t^2 =^32 16 t^2 = 2 t = 2 (t = – 2 is not a solution) t ≈ 1.41 The object will reach a hei ...
412 algebra De mystif ieD In the formula h = −16t^2 + v 0 t + h 0 , h 0 = 200 and v 0 = 0. h = −16t^2 + 200 We want to find t w ...
Chapter 11 QuaDraTiC appliCaTionS 413 Finally, we solve height problems for objects that are not dropped. We are given their ini ...
414 algebra De mystif ieD 1 4 16 60 44 1 4 ( tt^2 – + ) = () 0 4 t^2 – 15t + 11 = 0 (t − 1)(4t − 11) = 0 t t – = = 10 1 41 ...
Chapter 11 QuaDraTiC appliCaTionS 415 –+ 16 tt = 25 = 00 ( Thisiswhentheobjectisthhrown. –= – = – = = ) . 16 25 25 16 25 ...
416 algebra De mystif ieD The object will reach a height of 55 feet at about 0.35 seconds (on its way up) and again at about 0.9 ...
Chapter 11 QuaDraTiC appliCaTionS 417 Rectangle Formulas D L W FIGURE 11-1 • Area A = LW • Perimeter (the length around its ...
418 algebra De mystif ieD • Area of a circle A = πr^2 , where r is the circle’s radius. • Volume of a rectangular box V = LW ...
Chapter 11 QuaDraTiC appliCaTionS 419 A rectangle is 1 inch longer than it is wide. Its diameter is 5 inches. What are its dimen ...
420 algebra De mystif ieD The width of a rectangle is three-fourths its length. The diagonal is 10 inches. What are its dimensi ...
Chapter 11 QuaDraTiC appliCaTionS 421 The length is 7 inches more than the width, so L = W + 7. The diagonal is 17 inches. The ...
422 algebra De mystif ieD The classroom’s length is 14 feet more than its width, so L = W + 14. The diameter is 34 feet. The fo ...
Chapter 11 QuaDraTiC appliCaTionS 423 The hypotenuse of a right triangle is 34 feet. The sum of the lengths of the two legs is 4 ...
424 algebra De mystif ieD The volume of a box is 72 cm^3. Its height is 3 cm. Its length is 1.5 times its width. What are the le ...
Chapter 11 QuaDraTiC appliCaTionS 425 Let x represent the cup’s current radius. Then the radius of the new cup would be x + 1. T ...
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