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Mathematics
consider the quadratic functon y=-3(x-1)(x+2)
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Suppose you know that f(x) is an odd functon on the domain of all real numbers and that the function is concave up on the
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Nevermind, I figured it out. Since the function is odd it must be symmetric about the origin, so:
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A monic quadratic is a quadratic in which the coefficient of the quadratic term is 1. For example, r^2 - 3r + 7 is a monic
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check below http://www.jiskha.com/display.cgi?id=1433381900
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A monic quadratic is a quadratic in which the coefficient of the quadratic term is 1. For example, r^2 - 3r + 7 is a monic
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it is wrong!
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A monic quadratic is a quadratic in which the coefficient of the quadratic term is 1. For example, r^2 - 3r + 7 is a monic
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(x+16)(x-2) = x^2 + 14x - 32 (x+36)(x-2) = x^2 + 34x - 72 we know that +14 is correct, but -32 is
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Find the domain and range of the functon: f(x)=1/2 |x-2|
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domain is all real x. Any old x will do range--> well it will be zero when x = 2. If x is 4, it will
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Find the first partial derivative of the functon
f ( x , y ) = 2 x 3 y 2 + y 3
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assuming you meant f(x,y)=2x^3y^2+y^3 f<sub><sub>x</sub></sub> = 6x^2y^2 f<sub><sub>y</sub></sub> =
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The statement that is false is
A. A system of quadratic-quadratic equations can have exactly one solution. B. A system of
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C imagine a straight line and a parabola. How can they be the same graph?
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e^c-2c = e-2
what is c? ummm, 1? e^1 = e 2*1 = 2 No! Answer is 0.351 How do I get this? c = 1 is one solution. To find the other,
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To solve the equation e^c-2c = e-2, you can use Newton's method to find an approximate solution.
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consider the following quadratic equations: -y^2-16y-64=0 step 1 of 2: find the values of a,b, and c that should be used in the
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The given quadratic equation is -y^2 - 16y - 64 = 0. To find the values of a, b, and c for the
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consider the following quadratic equations: x^2-7x+2=0 step 1 of 2: find the values of a,b, and c that should be used in the
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The quadratic equation is in the form of ax^2 + bx + c = 0. To determine the values of a, b, and c
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