What happens to a 2nd degree equation when a changes?
93062
x^2 + x + a = 0 <--- This is an example of a second degree equation and I think you are asking how it would change with a?  If so the amount of curvature would change as the a changes.  (more open less open curve)
304526
let a-x=u\nso, -dx=du\n\nthen ∫1/(a-x)^2dx= \n-∫1/[(u)^2]du=\n-∫u^(-2)du=\n\n-[u^(-2+1)]/(-2+1)=\n\n-u^(-1)/-1=\nu^(-1)=\n\n1/u + C=\n\n1/(a-x) + C
309905
If you have a surface, from an equation of second degree with three unknowns (x^2, y^2, z^2), IF the coeficients are positive, the graph 'curves' to the inward... so, the surface closes. An example is a sphere. \n\nIF at least one of the coeficients is negative, the graph on that axis 'curves' to the outward, so it 'opens'...
1126609
Actually, no.\n\nIf dy/dx = 2, then I'd say yes, in moving from one coordinate pair to another, the y-value rises twice as fast as the x-value.  (That would be the case for a line with a slope of 2.)\n\nBut when dy/dx = 2x, as it does with a curve like y = x^2 + C, the ratio (of movement in the y-direction to movement in the x-direction) isn't constant -- it varies depending on where you are on the curve.  At the point (1, 2), for instance, that "slope," a.k.a. the slope of the tangent line at that point, is 2 -- the ratio of y-movement to x-movement is 2:1.  But at the point (3, 9), the slope of the tangent line is 6.\n\nHope that helps clear it up.  The important thing is that the slope of the tangent line is changing depending on where you are on the curve.
488850
Depress the cubic to eliminate the second order term.\n\nsee\nhttp://www.karlscalculus.org/cubic.html
940602
You're correct. Differentiating them and showing they are equal to zero will indicate that they are constant. The equation\n\nA(w_1)^2 + B(w_2)^2 + C(w_3)^2\n\ndifferentiated with respect to time yields\n\n2*A*w_1*(d/dt w_1) + 2*B*w_2*(d/dt w_2) + 2*C*w_3*(d/dt w_3)\n\nIf you substitute from the system of equations that shows the evolution then\n\n2*w_1*w_2*w_3*(B-C) + 2*w_1*w_2*w_3*(C-A) + 2*w_1*w_2*w_3*(A-B)\n\nyou can see that the terms cancel leaving you with zero.\n\nFollow the same technique for the second equation. It will look similar except you will wind up with an additional constantt, e.g., A*(B-C), etc.
799142
Hi Kitty,\nHere is what, I think, you need.\n\nExothermic, dH < 0, Entropy increases (dS > 0), Reaction is spontaneous and dG < 0\n\nExothermic, dH < 0, Entropy decreases (dS < 0), Reaction is spontaneous only at low temps, if |T dS| < |dH|\n\nEndothermic, dH > 0, Entropy increases (dS > 0), Reaction is spontaneous only  at high temps, if T dS > dH\n\nEndothermic, dH > 0, Entropy decreases (dS < 0), Reaction is not spontaneous, dG > 0
43939
is the above question in first degreee derivative??
59552
we have \nx^2 -84x+1764=0\nit's equation of second degree\nsolution x=42
134473
The answer is obviously (A). The bar may shrink a little, but it can't disappear completely, as the options B, C and D imply!!\nHence, by process of elimination, the answer is (A). :-)
196036
If you divide 2 to the second equation, you will notice that it is identical to the first one, so they are dependent to each other, and there are infinitely many solutions to this problem.\n\n^_^
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
193440
Let x be the side of a square\nThen, area of the square is, A=x^2\nThe rate of change of area w.r.t side is dA/dx =2x\nIn the problem another parameter is given, that is time.It is given in the problem that there is a constant change in the length of the side at the rate 0.4cm /sec.Here we have to notice that corresponding change in area is not uniform throughout.Therefore we can only find instantaneous change in area or accumulated area in the given periodor interval.That means change in area at a given instant of time or interval.\nFor example, let initial area be 0.Now, to find the area between the interval a and b we have to reintegrate. ie. integration of dA between the interval a and b =integration of 2x in the interval a & b.\nie. [A]a&b= [x^2]a&b=b^2-a^2 square units\nFor instance, change in area between the interval 9th and 10th seconds is =[x^2]a&b=b^2-a^2 square units\nGiven, 0.4units/sec,considering intial length and time as 0,we have, length of the side on the 9th second is a=9*0.4=3.6units and in the same way length on the 10th second is b= 4units\nAccordingly change in area(accumulated area) at the 10th second is =b^2-a^2\n=16-12.96=3.04 square units [this is an instataneous change]\nPlease note that, this change cannot be written as units^2 /sec
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
593998
When you have a polynomial, the coefficients are integers.  For example:\n\nax^2 + bx + c = 0\n\nThis is the general formula for a second degree polynomial (x is to the 2, hence second degree).  In this case, a, b, and c are the integers.\n\nAlso, if you add other terms, such as x^3, they would also be multiplied by integers.
785530
Answer D.\n\nThis is the Nernst equation at standard conditions.
291866
The same equation can be written as \ny = 2x^2-3x+2\n\nNow, for the new equation, let it be written in terms of the new shifted coordinates (X,Y). In the new system, what was the origin (0,0) earlier, now becomes (-1,-2). Hence the equations relating the two systems would be\nX = x - (-1) and Y = y - (-2)\nor X = x + 1 and Y = y + 2\n\nWritten in another way,\nx = X - 1 and y = Y - 2\n\nSubsituting in the main equation,\n\n(Y-2) = 2(X-1)^2 - 3(X-1) + 2\nOn simplifying,\nY = 2X^2 - 7X + 9\n\nHence, the general equation that you need is\nf1(x) = 2x^2 - 7x + 9