Assuming that you want (ab^3)^4, which is impossible to ask given the crap browser used by Answers, the solution is A^4b^12.
x-ab=0 x=ab
ab(ab) =2ab
There are 48 possibilities where 2 of 5 items must be adjacent.If you have 5 items, and they can be arranged in any order, there are 1205 x 4 x 3 x 2 ways to arrange them, for example ABCDE, ABCED, and so forth.However, if any two need to be placed next to each other, the number of variations is reduced to [4 x 3 x 2] x2 (=48), where there are only 4 separate "units" arranged, but the double-unit can appear with either of the pair first.For example, if A and B must be together, you have 24 possibilities:(AB)CDE(AB)CED(AB)DCE(AB)DEC(AB)EDC(AB)ECDC(AB)DEC(AB)EDD(AB)CED(AB)ECE(AB)CDE(AB)DCCD(AB)EDC(AB)ECE(AB)DEC(AB)DDE(AB)CED(AB)CCDE(AB)CED(AB)DEC(AB)DCE(AB)EDC(AB)ECD(AB)and another 24 where (AB) is replaced by (BA).
If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.
yes
Calculus AB is a Calculus course taught in high schools based on an AP curriculum. The class is supposed to ultimately prepare a student to take the AP Calculus AB exam in May. While the specifics might vary from school to school, the core of the curriculum are limit definitions, differentiations, integrations, and applications of all of the above.
Calculus AB is the branch of the calculus curriculum offered by the College Board exam administration. It differs from its counterpart, Calculus BC, in that AB only covers material into a standard "college" introductory calculus course.Usually, Calculus AB corresponds to a typical university's Calculus I course (or similarly named). In contrast, BC is a more accelerated course which covers material that corresponds to both Calculus I and II. This may vary depending on the university, as some combine both Calculus I and II into one course, in which case AB would only cover one semester of material.Calculus AB includes content based on limits, continuity, differentiation, integration, the IVT, the MVT, the EVT, and the FTC.
AP Calculus AB is a post-secondary course that is also offered in many high schools. Students that score high on the AP exam may not have to take certain college math courses.
Calculus AB tends to be easier due to it being mostly composed of the basic introductory to Calculus. Whereas Calculus BC contains information from AB but as well as further information which could possibly be somewhat more rigorous than AB.
AB C
Short answer: They're similar, but Calculus AB covers a bit more (and goes more in-depth) than Calculus 1. Long answer: The AP Calculus AB test covers differentiation (taking derivatives) and early integration (taking antiderivatives), including the concept/applications of an integral and integration by substitution. In college, Calculus 1 covers mostly differentiation and Calculus 2 covers additional strategies for integration and series. I like to think of it like this: A = Differentiation B = Integration C = Series So Calculus AB covers differentiation and integration and Calculus BC covers integration and series. College is more like: Calc 1 = A Calc 2 = B&C Of course, this depends on how much you cover in high school and college.
(a/b)'= (ba'-ab')/(b²)
I don't think they stand for anything. AB just means the first and second parts, and BC is the next parts.
Study every single thing in math
In High School, BC Calculus is an accelerated course in Calculus. It includes a full-year of the college course, while AB includes only half a year's worth. BC tends to move at a much faster pace and covers much more material than AB.
ab * -b = -ab^-2