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MrsDavidRandle

I need help!! Algebra II A Unit 5 Lesson 4!

  • one year ago
  • one year ago

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  1. MrsDavidRandle
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    1. Classify 3x5 – 8x3 – 2x2 + 5 by number of terms. (1 point) trinomial polynomial of 4 terms binomial polynomial of 5 terms Consider the leading term of the polynomial function. What is the end behavior of the graph? 4x5 + 1x 2. (1 point) Since n is odd and a is positive, the end behavior is up and down. Since n is odd and a is positive, the end behavior is down and up. Since n is odd and a is positive, the end behavior is down and down. Since n is odd and a is positive, the end behavior is up and up. 3. What is the graph of y =x3 (1 point) 4. Use synthetic division to find P(3) for P(x) = x4 – 6x3 – 4x2 – 6x – 2. (1 point) 3 –137 299 –47 5. What is a cubic polynomial function in standard form with zeros 1, –2, and 2? (1 point) = x3 + x2 – 3x + 4 = x3 + x2 – 4x – 2 = x3 + x2 + 4x + 4 = x3 – x2 – 4x + 4 6. What is the relative maximum and minimum of the function? = x3 + 6x2 – 36x (1 point) The relative maximum is at (–6, 216) and the relative minimum is at (2, –40). The relative maximum is at (–6, 40) and the relative minimum is at (2, –216). The relative maximum is at (6, 216) and the relative minimum is at (–2, –40). The relative maximum is at (6, 40) and the relative minimum is at (–2, –216). 7. Find the real solutions of the equation by graphing. x2 – x + 2 = 0 (1 point) x = 1 x = 3 No Solution x = 2 8. Over two summers, Ray saved $800.00 and $600.00. The polynomial represents his savings at the beginning of the third year, where x is the growth factor. (The interest rate r is x – 1.) What is the interest rate he needs to save $1,650.00 after the third summer? 800x2 + 600x (1 point) 0.1% 10.9% –285.9% 1.1% 9. Divide 3x3 + 3x2 + 2x – 2 by x + 3 using long division. (1 point) 3x2 – 6x + 20 3x2 + 12x – 16 3x2 + 12x – 16, R 58 3x2 – 6x + 20, R -62 10. Divide using synthetic division. –3x3 + 11x2 – x + 20 by (x – 4). (1 point) –3x2 + 23x + 3, R 40 –3x2 – x – 5 3x2 + x + 5 3x2 – 23x – 3, R –40 11. Miguel is designing shipping boxes that are rectangular prisms. The shape of one box, with height h in feet, has a volume defined by the function V(h) = h(h – 5)(h – 6). Graph the function. What is the maximum volume for the domain 0 < h < 6? Round to the nearest cubic foot. (1 point) 29 ft3 27 ft3 24 ft3 6 ft3

    • one year ago
  2. Trillography
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    1. Classify 3x5 – 8x3 – 2x2 + 5 by number of terms. (1 point) (0 pts) trinomial (1 pt) polynomial of 4 terms (0 pts) binomial (0 pts) polynomial of 5 terms 1 /1 point Consider the leading term of the polynomial function. What is the end behavior of the graph? 4x5 + 1x 2. (1 point) (0 pts) Since n is odd and a is positive, the end behavior is up and down. (1 pt) Since n is odd and a is positive, the end behavior is down and up. (0 pts) Since n is odd and a is positive, the end behavior is down and down. (0 pts) Since n is odd and a is positive, the end behavior is up and up. 1 /1 point 3. What is the graph of y =x3 (1 point) (1 pt) (0 pts) (0 pts) (0 pts) 1 /1 point 4. Use synthetic division to find P(3) for P(x) = x4 – 6x3 – 4x2 – 6x – 2. (1 point) (0 pts) 3 (1 pt) –137 (0 pts) 299 (0 pts) –47 1 /1 point 5. What is a cubic polynomial function in standard form with zeros 1, –2, and 2? (1 point) (0 pts) = x3 + x2 – 3x + 4 (0 pts) = x3 + x2 – 4x – 2 (0 pts) = x3 + x2 + 4x + 4 (1 pt) = x3 – x2 – 4x + 4 1 /1 point 6. What is the relative maximum and minimum of the function? = x3 + 6x2 – 36x (1 point) (1 pt) The relative maximum is at (–6, 216) and the relative minimum is at (2, –40). (0 pts) The relative maximum is at (–6, 40) and the relative minimum is at (2, –216). (0 pts) The relative maximum is at (6, 216) and the relative minimum is at (–2, –40). (0 pts) The relative maximum is at (6, 40) and the relative minimum is at (–2, –216). 1 /1 point 7. Find the real solutions of the equation by graphing. x2 – x + 2 = 0 (1 point) (0 pts) x = 1 (0 pts) x = 3 (1 pt) No Solution (0 pts) x = 2 1 /1 point 8. Over two summers, Ray saved $800.00 and $600.00. The polynomial represents his savings at the beginning of the third year, where x is the growth factor. (The interest rate r is x – 1.) What is the interest rate he needs to save $1,650.00 after the third summer? 800x2 + 600x (1 point) (0 pts) 0.1% (1 pt) 10.9% (0 pts) –285.9% (0 pts) 1.1% 1 /1 point 9. Divide 3x3 + 3x2 + 2x – 2 by x + 3 using long division. (1 point) (0 pts) 3x2 – 6x + 20 (0 pts) 3x2 + 12x – 16 (0 pts) 3x2 + 12x – 16, R 58 (1 pt) 3x2 – 6x + 20, R -62 1 /1 point 10. Divide using synthetic division. –3x3 + 11x2 – x + 20 by (x – 4). (1 point) (0 pts) –3x2 + 23x + 3, R 40 (1 pt) –3x2 – x – 5 (0 pts) 3x2 + x + 5 (0 pts) 3x2 – 23x – 3, R –40 1 /1 point 11. Miguel is designing shipping boxes that are rectangular prisms. The shape of one box, with height h in feet, has a volume defined by the function V(h) = h(h – 5)(h – 6). Graph the function. What is the maximum volume for the domain 0 < h < 6? Round to the nearest cubic foot. (1 point) (0 pts) 29 ft3 (0 pts) 27 ft3 (1 pt) 24 ft3 (0 pts) 6 ft3

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