New SAT Math Workbook

5 the correct answer is 21 first find r 3 r 31 147

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Unformatted text preview: hoice (B) holds for both pairs: y = 2x + 1 (−1) = 2(−1) + 1 (0) = 2(− 1 ) + 1 2 3. 8. = x 6 y4 yx 64 = x2 y2 (A) One point on l is defined by (–5,–6). A second point on l is defined by (4,0), which is the point of x-intercept. With two points defined, you can find the line’s slope (m) as follows: 62 2 1 m = x − x = 4 − (−5) = 9 = 3 . 2 1 y −y 0 − (−6) www.petersons.com Numbers and Operations, Algebra, and Functions 265 9. (C) The graph shows a parabola opening upward with vertex at (3,0). Of the five choices, only (A) and (C) provide equations that hold for the (x,y) pair (3,0). Eliminate choices (B), (D), and (E). In the equation given by choice (A), substituting any non-zero number for x yields a negative y-value. However, the graph shows no negative yvalues. Thus, you can eliminate choice (A), and the correct answer must be (C). Also, when a parabola extends upward, the coefficient of x2 in the equation must be positive. Exercise 1 1. (E). Solve for T in the general equation a · r (n – 1) = T. Let a = 1,500, r = 2, and n = 6 (the number of terms in the sequence that includes the value in 1950 and at every 12-year interval since then, up to and including the expected value in 2010). Solving for T: 1, 500 × 2(6−1) = T 1, 500 × 25 = T 1, 500 × 32 = T 48, 000 = T 10. (E) The following figure shows the graphs of the two equations: Doubling every 12 years, the land’s value will be $48,000 in 2010. 2. (A) Solve for T in the general equation a · r (n – 1) = T. Let a = 4, r = 2, and n = 9 (the number of terms in the sequence that includes the number of cells observable now as well as in 4, 8, 12, 16, 20, 24, 28, and 32 seconds). Solving for T: 4 × 2(9−1) = T 4 × 28 = T 4 × 256 = T 1, 024 = T 32 seconds from now, the number of observable cancer cells is 1,024. 3. (B) In the standard equation, let T = 448, r = 2, and n = 7. Solve for a : a × 2( 7−1) = 448 a × 26 = 448 a × 64 = 448 a = 448 64 a=7 As you can see, the graphs are not mirror images of each other about any of the axes described in answer choices (A) through (D). www.petersons.com 266...
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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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