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Exponentiation

Conic Sections, Rational Exponents, and Radical Functions Review

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Apcs solution acsl

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American Computer Science League Flyer Solutions 1. Boolean Algebra )( BA? CBBA ?( ) = BA CBBA ?( ) = 000 ???? CBBABAA 2. Computer Number Systems 18 = 1, 1002 = 4, 118 = 9, 1016 = 16 so the sequence is 1, 4, 9, 16 ? n2 . The 10th term is 102 = 100 = 1448 3. LISP (EXP (DIV (MULT (ADD 2 (SUB 4 2 ) ) 3 ) 2 ) 4 ) = (EXP (DIV (MULT (ADD 2 2 ) 3 ) 2 ) 4 ) = (EXP (DIV (MULT 4 3 ) 2 ) 4 ) = (EXP (DIV 12 2 ) 4 ) = (EXP 6 4 ) = 1296 4. Prefix/Infix/Postfix 4 3 + 7 5 - * 2 ^ = (4 + 3) (7 - 5) * 2 ^ = 7 * 2 2 ^ = 14 ^ 2 = 196 5. Bit String Flicking (LCIRC-3 (RSHIFT-2 X)) = 10001 Let X = abcde RSHIFT-2 abcde = 00abc LCIRC-3 00abc = bc00a bc00a = 10001 b = 1, c = 0, a = 1, d = * and e = *

LIMITS AND CONTINUITY

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Limits and Continuity Brief Review Limit ? intended height (y-value) of the function. Properties: add, subtract, divide, multiply, multiply constant and raise to any power. Techniques to Evaluation: Direct Substitution ? plug the x-value in?if you get a number you are done?if you get an indeterminate form?. 1.) Try to factor the expression. Cancel common factors and try direct substitution again. 2.) Try tables or graphs?.try plugging in a number close to the x-value to the right and the left. 2. 3. 4. 5. 6 7. 8.

Logarithms

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Many students in high school and in college have a difficult time with logarithms. In many cases, they memorize the rules without fully understanding them, and they sometimes even manage to squeak by a course. Why waste their time on these archaic entities; they are never going to see them again. Wrong! Just when the student breathes a sigh of relief to be done with logarithms, they encounter them again in another course. They are now in trouble because the second encounter with logarithms is at a more sophisticated level. Without an understanding of the basics, the student is doomed to blindly stumble through and fail the course. You have our sympathy and you have our solution. We at S.O.S. Math want you to succeed.

solving exponential equations

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to solve exponential equations without logarithms, you need to have equations with comparable exponential expressions on either side of the "equals" sign, so you can compare the powers and solve. In other words, you have to have "(some base) to (some power) equals (the same base) to (some other power)", where you set the two powers equal to each other, and solve the resulting equation. For example: Solve 5x = 53. Since the bases ("5" in each case) are the same, then the only way the two expressions could be equal is for the powers also to be the same. That is: x = 3

Exponents

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Since an exponent on a number indicates multiplication by that same number, an exponent on a negative number is simply the negative number multiplied by itself a certain number of times: (- 4)3 = - 4× -4× - 4 = - 64 (- 4)3 = - 64 is negative because there are 3 negative signs--see Multiplying Negatives. (- 5)2 = - 5× - 5 = 25 (- 5)2 = 25 is positive because there are 2 negative signs.

Inverse Hyperbolic Functions

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Since hyperbolic functions are defined in terms of exponential functions, it is expected that their inverses can be expressed in the inverse of their exponential functions: Inverse Hyperbolic Functions in terms of logarithms 

7inver1,7inver2,7inver3,7inver4,7inver5,7inver6,7inver7,7inver8,7inver9,7inver10,7inver11,7inver12



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