Math

Solve Y=-0.02x^2+1.3x+8

  1. 👍 0
  2. 👎 0
  3. 👁 141
asked by Elijah
  1. Assuming the question requires the zeroes of Y(x).

    Multiply by 50 (to get integer coefficients)
    50*y(x)=-x²+65x+400=0
    Solve by quadratic formula:
    x=(-65±√(65²+1600))/-2
    =5.66 or 70.66 approximately

    1. 👍 0
    2. 👎 0
    posted by MathMate

Respond to this Question

First Name

Your Response

Similar Questions

  1. ALGEBRA

    for subtracting this problem: (0.04x^3-0.03x^2+0.02x)-(0.03x^3+0.08x^2-6) I know to change the signs of the second grouping. When I done that, I cam up with: 0.04x^3-0.03x^2+0.02x-0.03x^3-0.08x^2+6 THEN .01x^3 - .11x^2 + 0.02x +

    asked by GGIFT on February 25, 2010
  2. Calculus

    Could someone work this out so I understand it. Thanks Given the supply function p=S(x)=5(e^0.02x-1) Find the average price (in dollars)over the supply interval [31,36] The 0.02x in the only thing ^ on the e

    asked by Jay on April 25, 2011
  3. Math

    Factoring / solve for x 5.50x = 0.02x^3 + 5x

    asked by Steph on November 28, 2012
  4. Algebra

    .04x + 720 - .06x > = 650 -.02x = 650 -.02x > = -70 x > = -70/-.02 x > = 3500 Which would be the solution to solve this inequality? .04 x + .06(12000-x) >= 650

    asked by Crystal on July 14, 2009
  5. Algebra

    Please solve the inequality .04 x + .06(12000-x) >= 650 .04x + 720 - .06x ¡Ý 650 -.02x ¡Ý -70 x ¡Ý -70/-.02 x ¡Ý 3500 I think this is wrong and what is below is correct? Which is right? Thanks! .04x + 720 - .06x ¡Ý 650

    asked by Crystal on July 14, 2009
  1. Math

    Could someone answer this question so I understand it. Thanks Given the supply function p = S(x)=5(e^0.02x - 1) find the average price(in dollars) over the supply interval [31,36] only the 0.02x is ^ on the e give the answer as an

    asked by Jules on April 27, 2011
  2. Math

    Can someone show me step by step on how to factor/solve this problem explaining the steps? 5.50 = (0.02x^3 + 5x) / x

    asked by Steph on November 29, 2012
  3. FINANCIAL MATHEMATICS

    The variable costs associated with a certain process are $0.65 per item. The fixed costs per day have been calculated as $200 with special costs estimated as $0.02X^2, where X is the size of the production run (that is, number of

    asked by FAKAAPO.SUETUSI on October 6, 2011
  4. math

    lim as x heads toward infinity of x^200/e^.02x

    asked by Anonymous on November 18, 2010
  5. math

    I need help in solving this subtraction (0.06x^3-0.07x^2+0.04x)-(0.02x^3+0.07x^2-9)

    asked by shelly on May 10, 2009

More Similar Questions