*According to dictionary.com, discriminant defines to: an expression used to distinguish or separate other expressions in a quantity or equation.\n\n*So the discriminant is basically the expression that is used to separate the other expressions from one another? How do you even determine that?\n\nFor example, what are the discriminants of these?\n\n3. x^-3x-4=0\n\n6. x^+2x+=0
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If you have a quadratic equation of the form ax squared + bx + c = 0, the discriminant is\nb squared minus 4 times a times c, or b^2 - 4ac\n\nIt's the part under the square root in the Quadratic Formula.  It tells if the roots  are real, rational, or imaginary.\n\nSo in your first example, it's 9 - 4X1X-4, or 9 - -16, or 25.
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Quick Definitions\n\nAll these definitions use the set of numbers  3, 5, 7, 3, 11, 1.\n\nAverages--\n\nMEAN\nThe mean is all of the numbers added up (3+5+7+3+11+1=30) divided by how many numbers there is (30/6) so the mean in this case is |5|\n\nMEDIAN\nTo find the median you have to arrange the numbers in order of size (1,3,3,5,7,11) and then it is the median is the middle number. If there is an even amount of numbers, then you take the mean of the two middle numbers (3+5=8; 8/2=4) so in this case the mean would be |4|\n\nMODE\nThe mode is simply the most common number. In this case 3 appears more than any of the other numbers, so 3 is the mode.\n\nHope this all helps with your math classes!
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Depends on what variable you're differentiating with respect to:\ndV/dr : 2/3*pi*r*h\ndV/dh:  1/3*pi*r^2
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that's as low as it'll go, nothing can be divided out of anything in the numerator or the denominator.  That's because there's not any like terms.  In other words, c nor d is in both parts of either the top or the bottom.  I hope you understood that, but trust me on this one.  Calculus us hard.
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subtracting a negative is the same as adding a positive\n\nmultiplying/dividing like signs (+,+ ; -,-) will give a positive product/quotient\n\nmultiplying/dividing opposite signs will give a negative answers
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linear operator takes a vector and maps it into another vector while preserving addition and multiplication by a scalar:\n\nL(ax+y)=aL(x)+L(y)\nx,y are vectors, a is a scalar, L() is operator\n\nIt can be defined by a matrix, i.e. result of operator is given by multiplying that matrix by a vector.\n\ngo look it up on wikipedia.org or mathworld.wolfram.com
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Okay, equations of the form ax^2+bx+c=0 have the discriminant b^2-4ac. In your case a=4,b=-3,c=5. So you have to take (-3)^2-4*4*5=-71. Since the discriminant is negative, you have 0 real solutions. If it were 0, you would have 1 real solution, and if it were positive, you would have 2 real solutions. Hope that helps!
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Assuming A, B, and C to be constants, this is an equation that defines a linear relation between variables x and y. There are a number of places where this would appear, but it would be easier to see it in the form y = -(A/B)x + (C/B), its equivalent.
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discriminant zero, x2 +/- 2x + 1\ndiscriminant +ve, x2 +/- 3x + 1\ndiscriminant -ve, x2 +/- x +1\n\nnow you take a graph paper and draw the three lines of three equaitions mentioned above using following 3 sets of locations\n\ndiscriminant zero, x2 + 2x + 1 \n(x,y) = (0, 1); (1, 4); (-1, 0)\nfor x2 - 2x +1\n(x,y) = (0,1); (1, 0); (-1, 4) \n\ndiscriminant +ve, x2 + 3x + 1\n(x, y) = (0, 1); (1, 5); (-1, -1)\nfor x2 - 3x +1\n(x,y) = (0, 1); (1, -1); (-1, 5)\n\ndiscriminant -ve, x2 + x +1\n(x,y) = (0, 1); (1, 3); (-1, 1)\nfor x2 - x +1\n(x,y) = (0, 1); (1, -1); (-1, 3)\n\nHope this come to your help
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Assuming by "plus" and "minus", you are referring to "positive" and "negative"... they are fundamental rules for mathematics.  To understand them, you need to understand the definition of a "negative number", where subtraction of a positive number is the equivalent of addition of the number's negative form:\n\n5 - 3 = 5 + (-3) = 2\n\nYou can think of negative values as the "explicit absence of" a value or number of items.  With this definition, it should be clear why the multiplicative rules are as you stated.\n\n(positive) * (positive) = (positive) ...is just the obvious, since any positive number of items added multiple times will always result in a positive number of items.  Add 3 apples to a basket 4 times, and you will have 3*4=12 apples.  Mathematically:\n\n3 + 3 + 3 + 3 = 12\n\n(positive) * (negative) = (negative) ...you can use the "explicit absence of" idea I stated above to visualize this one.  If you have 2 holes into which a ball would fit and consider that number to be -2 (negative 2), and you had three rows of them, you will have 3 * (-2) = -6 holes, into which 6 balls would fit.  Mathematically:\n\n(-2) + (-2) + (-2) = -6\n\n(negative) * (negative) = (positive)  ...is derived directly from the 2nd rule above.  If a single negative number changes the sign of the product, then two of them should change the sign right back.  You can also use the "explicit absence of" idea.  If you were counting pairs of holes as negative values (-2), but you were to fill up 4 of these pairs with balls (the filling of which would be considered as "negating the hole", or creating a filled hole), then you would have (-2) * (-4) = 8 holes filled with balls.  Mathematically:\n\n-[ (-2) + (-2) + (-2) + (-2) ] = -(-8) = 8\n\nI hope this makes some sense, and that I didn't just create even more confusion.  ^.^
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Because subtraction is defined as the inverse of addition.\nThis means that when a+b =c (addition),  subtraction means\n c-b=a  or c-a=b.  And conversely, when  c-b=a, then a+b=c.\nBy the way, that's why we add the subtracter and the difference\nto see if we get the subtrahend when we check subtraction.\n\n  Now suppose we say  c-(-b)=a.  New scenario. We know \nthat a+(-b) = c which is equivalent to a-b=c. And this means\nthat  c+b =a.  Hey, did you catch that. I just went from c-(-b)=a\nto c+b=a and doesn't that show you that subtracting -b is the same as adding +b.
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The discriminant = b^2-4ac = 1 - 4.2.(-4) = 33 > 0 \n⇒ We can get two unequal, real but irrational roots.\n\nThis is so, because, by actually calculating the roots of the equation applying formula. we find that:\nx= [1±√33]/4
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The first answer is not exactly right. The roots of the quadratic are\n(-b(+/-)sqrt(b^2-4*a*c)) / (2*a) (not -b^2...) \nHowever the solution is correct; the discriminant is b^2-4*a*c which for the first quadratic is 1^2 - 4*5*2 = -39. Per the ref., a negative discriminant means both roots are complex. They are: (-1(+/-)sqrt(39)*i ) / 10.\nFor the second quadratic, the discriminant is 4*4 - 4*4*1 = 0. As a result there is one real root: -4 / 8 or -0.5.\nRe your followup question, the summary is:\n1st equation, disc. = -39, there are no real roots and 2 complex roots, \n(-1+sqrt(39)*i ) / 10 and (-1-sqrt(39)*i ) / 10.\n2nd equation, disc. = 0, there is one real root (0.5) and no complex roots.
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basically, think of a function as a machine:\n\ntake this an an example: f(x) = 2x +3\nsee the x in brackets?\n\nif x = 1, then f(1) = 2(1) + 3 = 5\nif x = 2, then f(2) = 2(2) + 3 = 7\nif x = 3, then f(3) = 2(3) + 3 = 9\n\nsee whats happening? the answer DEPENDS on number inputted in the brackets.\n\nso therefore, the definition you quoted "a consistent, predictable, relationship between the independant and dependant variable" is true.
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I'm not certain what the expression is.\nI'm getting 2 + [y²/(x^4 + x^3*y)]\n\nI assumed that the total denominator is the x²+xy.  So I'm ignoring that for now.\n\nThen the total numerator has a least common denominator of x².  So I got (2x^4 + 2x^3*y+y^2) / x².  Now go back and multiply all that by 1/(x²+xy)\nthe denominator is x^4+x^3*y.  You see the first two terms in the numerator?  they are this new denominator *2.  So we can move them out and have 2+ the remainder [y² / (x^4+x^3*y)]\n\n-edit-\n\nThen put answer D.  You don't understand how to phrase math, so none of us can understand the question.  However, (x+y) is definitely a factor in the denominator and D is the only answer that has it.
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its called the "discriminant" and it can be found here:
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Another way to look at is follows.\n\nThink of d/dx () to mean "differentiate whatever is in the parens with respect to x.  So, for example d/dx (x)=dx/dx=1.   If y=y(x). y'(x)=d/dx (y(x))=dy/dx where the argument of y is omitted for convenience.\n\nSo, d/dx (y')=d/dx (dy/ dx)=(d d y)/(dx dx)=d^2y/dx^2. \n\nThe notation is ambiguous.  One would usually think that the x is being square in the denominator, but "dx" is a single symbol, so in fact, in this context, dx^2=(dx)^2.\n\nI hope this helps!\n\nGood luck. :)