Review 5.1 to 5.3 Practice for Quiz.

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Presentation transcript:

Review 5.1 to 5.3 Practice for Quiz

Lesson Quiz: Part I Variation 1. The volume V of a pyramid varies jointly as the area of the base B and the height h, and V = 24 ft3 when B = 12 ft2 and h = 6 ft. Find B when V = 54 ft3 and h = 9 ft. 18 ft2 2. The cost per person c of chartering a tour bus varies inversely as the number of passengers n. If it costs $22.50 per person to charter a bus for 20 passengers, how much will it cost per person to charter a bus for 36 passengers? $12.50

Example 3 Given: y varies inversely as x, and y = 4 when x = 10. Write and graph the inverse variation function. k x y = y varies inversely as x. k 10 Substitute 4 for y and 10 for x. 4 = k = 40 Solve for k. 40 x Write the variation formula. y =

Example 3 Continued To graph, make a table of values for both positive and negative values of x. Plot the points, and connect them with two smooth curves. Because division by 0 is undefined, the function is undefined when x = 0. x y –2 –20 –4 –10 –6 –20/3 –8 –5 x y 2 20 4 10 6 20/3 8 5

5.2 Simplifying Rational Expressions Example 1 Simplify. Identify any x-values for which the expression is undefined. 6x2 + 7x + 2 6x2 – 5x – 5 (2x + 1)(3x + 2) (3x + 2)(2x – 3) (2x + 1) (2x – 3) Factor; then divide out common factors. = The expression is undefined at ????

Example 2: Multiplying Rational Expressions Multiply. Assume that all expressions are defined. 3x5y3 2x3y7  10x3y4 9x2y5 x – 3 4x + 20  x + 5 x2 – 9 A. B. 3x5y3 2x3y7  10x3y4 9x2y5 3 5 x – 3 4(x + 5)  x + 5 (x – 3)(x + 3) 3 5x3 3y5 1 4(x + 3)

Dividing: Example Divide. Assume that all expressions are defined. 2x2 – 7x – 4 x2 – 9 ÷ 4x2– 1 8x2 – 28x +12 2x2 – 7x – 4 x2 – 9  8x2 – 28x +12 4x2– 1 (2x + 1)(x – 4) (x + 3)(x – 3)  4(2x2 – 7x + 3) (2x + 1)(2x – 1) (2x + 1)(x – 4) (x + 3)(x – 3)  4(2x – 1)(x – 3) (2x + 1)(2x – 1) 4(x – 4) (x +3)

Example 5: Solving Simple Rational Equations Solve. Check your solution. x2 – 3x – 10 x – 2 = 7 (x + 5)(x – 2) (x – 2) = 7 Note that x ≠ 2. x + 5 = 7 x = 2 Because the left side of the original equation is undefined when x = 2, there is no solution.