a negative rational number
A strongly basic solution
A Weakly Acidic Solution
No because it's not a simultaneous equation but some kind of algebraic expression
it cant a negative + negative always equals a positive -Posted By Anonymous 7th Grader it cant be a negative plus a negative equals a negative always -Posted BY Anonymous 7th Grader♥
If the discriminant is positive, then the function has two real zeros. If it is zero, then the function has one real zero. If it is negative, then it has two complex conjugate zeros.This assumes that we are talking about a standard second order polynomial equation, i.e. quadratic equation, in the form Ax2 + Bx + C = 0, and that the discriminant is B2 - 4AC, which is a part of the standard solution of these kind of equations.
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I understand why you ask...but the answer is no.As a mathematical object, zero is an integer -- a number. As such, zero could easily be a SOLUTION to an EQUATION. But it's not the same as an equation. An equation is a different kind of mathematical object entirely.It's probably a good idea to try looking at zero not as "nothing", but as something: the middle point between positive and negative numbers.
A linear equation in one variable has one solution. An equation of another kind may have none, one, or more - including infinitely many - solutions.
There is none because its not an equation but some kind of algebraic expression.
H2O is the chemical equation (not solution) for water. It means: 2 atoms of hydrogen, and 1 atom of oxygen.
when you're really good in that kind of equation, but cannot, no matter what u do, CANNOT figure it out! when you look it up and can't find the answer in the back of the book. It depends on what kind of problem it is.
Salts produce (metallic) positive catIONS and negative anIONS (of non-metals)
The individual particles of an ionic crystal are positive and negative ions.
You can add or subtract any quantity on both sides of an equation, without changing the equation's solution set. Just make sure you add or subtract the same thing on both sides.
No kind: it is an expression, not an equation.
x² - 6x -8 = -56 <=> x² -6x + 48 = 0 We now calculate the discriminant (which equals b²-4ac for an equation of the form ax² + bx + c-: D = b²-4ac = (-6)² - 4*1*48 = 36 - 192 = -156 I don't know what kind of course in calculus you are taking, but if you only want the real answers to this equation, you can stop here because the discriminant is negative, meaning there are no real solutions. However, there are complex solutions to this equation The complex roots of D are sqrt(156)*i=12,5*i and -sqrt(156)*i=-12,5*i There are two solutions to a quadratic equation, namely: x1 = (-b + sqrt(D))/(2*a) and x2 = (-b - sqrt(D))/(2*a) so the two solutions we find for this equation are: x1 = (-b + sqrt(D))/(2*a) = (6+12,5i)/2 = 3+6,25i and x2 = (-b - sqrt(D))/(2*a) = (6-12,5i)/2 = 3-6,25i x1 and x2 are complex solutions to this quadratic equation.