# 501 Calculus Questions by Mark McKibben By Mark McKibben

Designed for these looking aid learning calculus at school - additionally helpful for adults trying to learn/re-learn calculus. A source for teachers supplementing their guide. 501 Calculus difficulties is helping clients arrange for educational checks and construct problem-solving abilities. in contrast to textbooks, complete solution factors are supplied for all difficulties. contains: - Questions protecting all of unmarried variable calculus - universal calculus errors - certain answers/fully labored suggestions - word list and theorem checklist - entry to loose on-line try

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Solve for r: V = 4 πr 2 h , assuming that r > 0 34. Solve for w: P = 2l + 2w 35. qxd 4/25/12 12:39 PM Page 16 501 Calculus Questions Answers 1. Apply the usual order of operations as follows: –4(3a2 – 24) = –4[3(–3)2 – 24] = –4[3(9) – 24] = –4[27 – 24] = –4 = –12 2. , those terms with the same variable) as follows: 4 ( k + m ) – 9 (3k – 2m ) + k + m = 4k + 4m − 27k + 18m + k + m = (4 − 27 + 1)k + (4 + 18 + 1)m = −22k + 23m 3. Apply the distributive law of multiplication, together with the usual order of operations, and gather like terms as follows: −3 {2a(−4a + 3) + 6(4 − a)} + 5a 2 = −3 {−8a 2 + 6a + 24 − 6a} + 5a 2 = = −3 {−8a 2 + 24} + 5a 2 = = 24a 2 − 72 + 5a 2 = 29a 2 − 72 4.

You must identify all possible y-values that are attained within the graph of f. The graph of f is comprised of three distinct components, each of which contributes an interval of values to the range of f. The set of y-values corresponding to the bottommost segment is (–2,1]; note that –2 is excluded due to the open hole at (5,–2) on the graph, and there is no other x-value in [–5,5] whose functional value is –2. Next, the portion of the range corresponding to the middle segment is [0,2); note that 2 is excluded from the range for the same reason –2 is excluded.

Qxd 4/25/12 12:39 PM Page 25 501 Calculus Questions 31. The strategy behind completing the square is to rewrite a quadratic equation of the form ax 2 + bx + c = 0 as an equivalent equation of the form a(x + e)2 + f = 0 , where a, b, c, e, and f are real numbers and a ≠ 0 . The reason is that the latter equation is easily solved using radical methods. In order to make this transformation, we must determine which real number w can be added to the expression x 2 + 12 x so that x 2 + 12 x + w factors as a perfect square (x + e)2 .