New SAT Math Workbook

For example substituting 0 1 and 1 for y gives us the

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Unformatted text preview: graph of y = , 3 except translated three units to the left. To confirm this, again, substitute simple values for x and solve for y in each case. For example, substituting 0, –3, and –6 for x gives us the three (x,y) pairs (0,3), (–3,0), and (–6,–3). Plotting these three points on the xy-plane, then connecting them with a curved line, suffices to show the same parabola as the previous one, except with vertex (–3,0) instead of (0,0). The equation | x | = y 2 represents the 1 union of the two equations x = y 2 and (D) −x = 1 y2 1 Plug either (x,y) pair into the standard equation y = mx + b to define the equation of the line. Using the pair (2,3): y = −x + b 3 = −2 + b 5=b 2. The line’s equation is y = –x + 5. To determine which of the five answer choices provides a point that also lies on this line, plug in the value –101 (as provided in the question) for x: y = –(–101) + 5 = 101 + 5 = 106. 3. x2 4. . The graph of the former equation is the hyperbola shown to the right of the y-axis in the figure, while the graph of the latter equation is the hyperbola shown to the left of the y-axis in the figure. www.petersons.com Numbers and Operations, Algebra, and Functions 271 5. (B) In this problem, S is a function of P. The problem provides three (P,S) number pairs that satisfy the function: (1, 48,000), (2, 12,000) and (4, 3,000). For each of the answer choices, plug each of these three (P,S) pairs in the equation given. Only the equation given in choice (B) holds for all three (P,S) pairs: = 48, 000 (1)2 48, 000 48, 000 = 12, 000 12, 000 = = 4 (2)2 48, 000 48, 000 3, 000 = = = 3, 000 16 (4) 2 48, 000 = 48, 000 Retest 1. The correct answer is 432. First, find r: 2 × r (3−1) = 72 2 × r 2 = 72 r 2 = 36 r=6 To find the fourth term in the sequence, solve for T in the standard equation (let r = 6 and n = 4): 2 × 6( 4−1) = T 2 × 63 = T 2 × 216 = T 432 = T 2. (D) The set of positive integers divisible by 4 includes all multiples of 4: 4, 8, 12, 16, . . . . The set of positive integers divisible by 6 includes all multiples of 6: 6, 12, 18, 24, . . . . The least common multiple of 4 and 6 is 12. Thus, common to the two set...
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This note was uploaded on 08/15/2010 for the course MATH a4d4 taught by Professor Colon during the Spring '10 term at Embry-Riddle FL/AZ.

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