MATH 110 SAMPLE QUESTIONS for Exam 1 on 9/9/05. 4x 2 (yz) x 4 y

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1 MATH 110 SAMPLE QUESTIONS for Exam 1 on //05 1. (R. p7 #) After simplifying, the numerator of x x z yz 1 4 4x (yz) 1 3 x 4 y. (R.3 #11) The lengths of the legs of a right triangle are 7 and 4. Find the length of the hypotenuse (R.4) Perform the indicated operation and simplify: (x 5)(3x + 4) 6x 0 5x x 0 6x 7x 0 5x 1 5x 7x 0 4. (R.4) Perform the indicated operation and simplify: (x ) 3 (4x 3 ) 3x 6x 8 3x 8 8x 8 5. (R.4) Simplify: (x 3x + 1) (x 5) x x 17x + 5 x 5x + 6 x x 4 x x + 6 x 5x 4 6. (R.4 Example ) (x + ) 3 = x x 3 6x + 1x 8 x 3 + 3x + 3x + 8 x 3 + 6x + 1x (R.4 Example b) Multiply out: (x 1) 3 x 3 1 x 3 3x + 3x 1 x 3 x x + 1 x 3 3x x (R.4 #64) Multiply and simplify: (x 3y)( x + y) x 5xy 3y x 7xy + 3y x 5xy + 3y x 7xy 3y. (R.4 #84) Expand and simplify: (x + 3y) x + 3y 4x + 6xy + y 4x + y 4x + 1xy + y

2 10. (R.5 Example 11) Factor the following expression completely: x x 1 (x 6)(x + ) (x 4)(x + 3) (x + 4)(x 3) (x + 6)(x ) 11. (R.5 #111) Factor completely: 3(x + 10x + 5) 4(x + 5) 3x + 6x + 5 (x + 5) (x + 5)(3x 1) (x + 5)(3x + 11) 1. (R.5 #1) Factor completely: 4(x + 5) 3 (x 1) + (x + 5) 4 (x 1) 8(x + 5) 7 (x 1) 3 (x + 5) 3 (x 1) 6(x + 5) 3 (x 1)(x + 1) (x + 5) 3 (x 1)(3x + 7) 13. (R.6) What the remainder when x + x + 1 divided by x? (R.6 #) When 5x 4 3x + x + 1 divided by x + the remainder x + 7 x + 13 x + 5 x (R.6 #10) When 5x 4 x + x divided by x + we get quotient: 5x 11; remainder: x 4 quotient: 5x 11; remainder: x + 0 quotient: 5x + 11; remainder: x 4 quotient: 5x + 11; remainder: x (R.6 #13) When x 4 3x 3 + x + 1 divided by x 3 1 the quotient x x x + x 1 x + x x x + 1

3 17. (R.7) Reduce the following rational expression to lowest terms: x + 3 x + x + 3 x x 6 4 x 18. (R.7) Simplify; leave your answer in factored form: 1 x + 1 x + 1 (x + 1)(x + ) 1 x 1 x + x + 4 (x 1)(x + ) x + x 6 x (R.7 #45) After simplifying, the numerator of x 1 x + 8 x + 6 x 8 ( ) x 0. (R.7) Simplify: x + 1 ( ) x + x x x + 4 x 3 x + 4x + x 3 x + 1. (R.7 #66) After simplifying x (x + ) (x 1) 6, the numerator (x + )(x 1) 8x 14 4x 14 8x x x (R.7 #7) After simplifying, the numerator of [ 1 1 h (x + h) 1 ] x x h x + h x + h x h

4 3. (R.7 #76) After simplifying, the denominator of x + 1 (x 1) 1 x x + 1 x 1 x (x + 1) x 1 4. (R.7) Simplify and factor x 5 1 x x + 1 (x + 5)(x 5) (x + 1)(x 1) (5x + 1)(5x 1) (x + )(x ) (x + 5)(x 5) 4x + 1 (5x + 1)(5x 1) 5. (R.8) Simplify: (R.8) Simplify: (R.8 Example 7) Simplify: (x /3 y)(x y) 1/ : x 8/3 y y3/ y3/ x /3 x 4/3 x 1/3 ( x y 1/3 ) 1/ 8. (R.8 Example 7c) Simplify: x 1/3 y 3x x /5 y 1/3 3x /5 y 1/3 3x 5/6 y 1/3. (R.8 #8) Simplify:

5 30. (R.8) Rationalize the denominator: (R.8) Simplify 4 4 ( ) / (R.8) Factor the expression x 1/ (x + x) + x 3/ 4x 1/ (where x 0). x 1/ (x + 1)(x 3) x 1/ (x + 6)(x 4) x 3/ (x + 1)(x 3) (x + 6)(x 4) x 3/ (x + )(x 6) 33. (R.8) Multiply and simplify ( x 3)( x + 5) 4x + 4 x 15 4 x 15 x + 4 x 15 4x (R.8 #51) After rationalizing the denominator of x + h x x + h + x, the numerator x + h x x(x + h) x + h x(x + h) x + h + x(x + h) 35. (1.1) Solve for x: 7 x = + 3x x = x = 5 x = x = 3 x = (1.1 Example 6) The solution to the equation 3x x 1 + = 3 x 1 x = 1 There no solution x = 5 x = 1 5

6 37. (1.1) Solve the equation: 1 1 x = 6 + x. x = 3 x = 10 3 x = x = (1.1 #4) Solve the equation x x + 3 = x. x = There no solution x = 1 x = 1 3. (1.1 #51) Solve th equation: x x 4 = 4 x 4 3 x + x = or x = There NO SOLUTION x = 1 x = 40. (see 1.1 #8) Going into the final exam, which will count as two tests, Brooke has test scores of 80,83,71,61, and 8. What score does Brooke need on the final in order to have an average score of 80? (1.1 #6) A wool suit, dcounted by 30% for a clearance sale, has a price tag of $3. What was the suit s original price? Not enought information to determine $53 $306. (approximately) $570 $ (1.) Solve for x: x 3x + = 0 x = 1 or x = x = or x = 3 x = 1 or x = x = or x = (1.) Solve for x: x x 4 = 0 x = 4 or x = x = or x = 1 5 x = x = + 5 or x = 5 or x = (1.) Find the value of a so that x + ax + 1 a perfect square. a = 1 3 a = 3 a = 1 a =

7 45. (1.) Find the value of k so that x 3 x + k a perfect square (1. #51) Use the quadratic formula to solve the equation x = 1 and x = x = 1 and x = x = 1 and x = 3 x = 1 and x = 3 3 x 5 x + 1 = 0. The solutions are (1. #83) Use the quadratic formula to find the real solutions: x + x 1 = 0. x = ± x = ± ± 3 x = x = ± (1. #7) An open box to be constructed from a square piece of sheet metal by removing a square of side 1 foot from each corner and turning up the edges. If the box to hold 4 cubic feet, then the dimensions of the sheet metal should be 1 foot by 1 foot 8 feet by 8 feet feet by feet 4 feet by 4 feet 4. (1. #) A ball thrown vertically upward from the top of a building 6 feet tall with an initial velocity of 80 feet per second. The height s (in feet) of the ball above the ground after t seconds s = t 16t. After how many seconds does the ball strike the ground? That, after how many seconds the height equal to zero? t = 1 t = t = 3 t = (1.) The equation 1 1 x 1 x = 0 NO real solutions has exactly ONE real solution, which POSITIVE exactly ONE real solution, which NEGATIVE exactly TWO real solutions, whose product 1 exactly TWO real solutions, whose product 1 1