y(y-3)(y+3)
Dividend: x3+4x2-9x-36 Divisor: x+3 Quotient: x2+x-12
Your answer will depend on the parameters of the instructions. If you're looking for the first derivative, simply use the product rule by changing the denominator to a negative exponent and bringing it up (take the negative square root of the quantity x-2 to the top). Then, follow the rules of calculus and algebra. Wow, that's a mess. Let's see... you get "the quantity x cubed plus 6x squared plus 3x plus 1 times the quantity -1(x-2) raised to the negative second plus the quantity x-2 raised to the negative first times the quantity 3x squared plus 12x plus 3." This is because of the Product Rule. Simplifying (by factoring out (x-2) raised to the negative second and combining like terms) gives us "(x-2) raised to the negative second times the quantity 2x cubed minus 24x minus 7." This can also be written as "2x cubed minus 24x minus 7 all over the quantity x-2 squared." f'(x)= 2x^3-24x-7 (x-2)^2
x over x is one, so the problem would be 1-2-3/2=-5/3
Your question is ambiguous. (3x)^3 times (3x)^3 = 27x^3 times 27x^3 = 729x^6 or, if you meant 3x^3 times 3x^3 = 9x^6
9 minus 8
How about: (3 squared minus 2 cubed) + 1 to the fourth 3 cubed minus 5 squared log 100 Kim Basinger's weeks minus Doris Day's cents.
(8.375 * 10^-3) = 0.008375, this cubed = 0.0000005874 or 5.874 * 10^-7
14^3 - 28 = 2716
y(y-3)(y+3)
xcubed-1 Answer::(X-1)(Xsquared+X+1) when you factor xcubed minus a number its the same thing as x cubed minus y cubed and x cubed minus y cubed factors to:: (x-y)(xsquared+xy+y squared) the first factor, (x-y), is the cubed root of the first and the cubed root of the second, so in the answer i have (x-1), which is x cubed minus one cubed :) the second factor, (xsquared+xy+ysquared), you take the first number squared, Xsquared, then the first and second one multiplied together, XY, and then the second number squared, Ysquared, so in the answer i have (xsquared+x+1), which is x squared, then x times 1 which is just x, and positive 1, which is negative 1 squared :) x^3 - 1
24
x(x + 3)(x - 4)
Locate the turning point(s) for the following functions. (a) y=x3-x2 -3x + 5 3
4x2 + 6x - 3 (with no remainder)
x^3 - x^3 = 0 Remember , whilst 'x' is an unknown value, that unkonwn is a fixed value. As a numerical example 3^(3) - 3^(3) = 27 - 27 = 0 The '3' is 'x' in this case
x³ + x² - 3x - 3 = (x + 1)(x + √3) (x - √3).