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maximum value of the carry
-3
-2x2 + 4x - 3 when x = -3 is -2*(-3)2 + 4*(-3) - 3 = -2*9 + 4*(-3) - 3 = -18 -12 -3 = -33
d/dx 2x2+3x+7=4x+3
2x2-4x+5 divided by x-1 Quotient: 2x-2 Remainder: 3
maximum value of the carry
-3
2x2-4x-3x+6 2x2-7x+6 (2x+3)(x+2)
-2x2 + 4x - 3 when x = -3 is -2*(-3)2 + 4*(-3) - 3 = -2*9 + 4*(-3) - 3 = -18 -12 -3 = -33
d/dx 2x2+3x+7=4x+3
2x2-4x+5 divided by x-1 Quotient: 2x-2 Remainder: 3
You can work this out with long division, by checking to see if (x2 - 1) is a factor of (2x4 + 4x3 - x2 + 4x - 3). It is. Unfortunately, the WikiAnswers system is somewhat limited in depicting things such as long division, so we won't be able to represent it here. In short though, (2x4 + 4x3 - x2 + 4x - 3) / (x2 + 1) is equal to 2x2 + 4x - 3. which means that: (x2 + 1) / (2x4 + 4x3 - x2 + 4x - 3) = (x2 + 1) / (x2 + 1)(2x2 + 4x - 3) = 1 / (2x2 + 4x - 3)
2(x^2 +2x + 3)
y = -1 + 3 sin 4xLet's look at the equation of y = 3 sin 4x, which is of the form y = A sin Bx, wherethe amplitude = |A|, and the period = (2pi)/B.So that the amplitude of the graph of y = 3 sin 4x is |3| = 3, which tell us that the maximum value of y is 3 and the minimum value is -3, and the period is (2pi)/4 = pi/2, which tell us that each cycle is completed in pi/2 radians.The graph of y = -1 + 3 sin 4x has the same amplitude and period as y = 3 sin 4x, and translates the graph of y = 3 sin 4x one unit down, so that the maximum value of y becomes 2 and the minimum value becomes -4.
If x = 3, 4x = 12
f(x) = 7 + 4x - 2x2Difference QuotientYou are going to plug in f(x) into:f '(x) lim Δx -> 0 = f(x + Δx) - f(x) / Δx[7 + 4(x + Δx) - 2(x + Δx)2] - (7 + 4x - 2x2) / Δx= 7 + 4x + 4Δx - 2x2 - 4xΔx - 4(Δx)2 - 7 - 4x + 2x2 / Δx= 4Δx - 4xΔx - 4(Δx)2 / Δx= Δx (4 - 4x - 4Δx) / Δx= (4 - 4x - 4Δx)= 4 - 4x - 4(0)= 4 - 4xYou can check using Power Rule:f(x) = 7 + 4x - 2x2 f'(x) = 0 + 4(1)x1-1 - 2(2)x2-1= 4 - 4xWhen x = 3:f'(3) = 4 - 4(3)= 4 - 12= -8
y = 2(x - 1)2 + 3 y = 2(x - 1)(x - 1) + 3 y = 2(x2 - 2x + 1) + 3 y = 2x2 - 4x + 2 + 3 y = 2x2 - 4x + 5