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New SAT Math Workbook Episode 2 part 6

Tham khảo tài liệu 'new sat math workbook episode 2 part 6', ngoại ngữ, ngữ pháp tiếng anh phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 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 y-values. 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. 10. E The following figure shows the graphs of the two equations 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 X 2 6-1 T 1 500 X 25 T 1 500 X 32 T 48 000 T 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 X 2 9-1 T 4 X 28 T 4 X 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 X 2 7-1 448 a X 26 448 a X 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 Chapter 15 4. The correct answer is 800. Solve for a in the general equation a r n - 1 T. Let T 2 700. The value at the date of the purchase is the first term in the sequence and so the value three years later is the fourth term accordingly n 4. Given that painting s value increased by 50 or 2 per year on average r 1.5 I Solving for a 2 700 a X 2 700 a X 27 2 700 8 a 2 700 X-8 27 a 800 At an increase of 50 per year the .