Math 1314 Test 2 Summer 2020 Practice With Solutions

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Mathematics

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Feb 20, 2024

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Math 1314 Practice Test 2 Solutions NAME____________________ Question Points Possible 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 Total 100 Make sure to include all work you want graded on your submission Must Show Work to get full credit.
1. Solve the following linear equation: 6 x + 7 = 4 x + 55 Answer: 6 x + 7 4 x 7 = 4 x + 55 4 x 7 2 x = 4 8 x = 24 2. The following rational equation has denominators that contain variables. For the equation 1 x + 3 + 2 = 9 x + 3 a. Write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. Answer: x + 3 = 0 x =− 3 b. Keeping the restrictions in mind, solve the equation. Answer: ( 1 x + 3 + 2 ) ( x + 3 )= ( 9 x + 3 ) ( x + 3 ) 1 + 2 ( x + 3 ) = 9 1 + 2 x + 6 = 9 2 x = 2 x = 1 3. The average cost of tuition and fees at public four-year colleges in the United States can be modelled by the following formula, where T represents the average cost of tuition and fees for the school year ending x years after 1996. When will the tuition and fees at public US college’s average $5411? T = 165 x + 2771 Answer: 5411 = 165 x + 2771 2640 = 165 x x = 16 16 + 1996 = 2012 They will average $5411 in 2012 4. Multiply and write the expression in standard form a + bi .
( 8 + 7 i ) ( 8 7 i ) Answer: 64 56 i + 56 i 49 i 2 = 64 + 49 = 113 5. To solve x 2 2 3 x = 4 9 by completing the square what must be added to both sides of the equation? Answer: ( 2 3 2 ) 2 = ( 1 3 ) 2 = 1 9 6. Find the x-intercepts of the graph of y = x 2 2 x 35 . Answer: x 2 2 x 35 = 0 ( x 7 ) ( x + 5 ) = 0 x = 7 , 5 7. Solve the quadratic equation x 2 2 x = 4 . Make sure you solutions are simplified. Answer: ( 2 2 ) 2 = ( 1 ) 2 = 1 x 2 2 x + 1 = 4 + 1 ( x 1 ) 2 = 5 ( x 1 ) 2 = ± 5 x 1 = ± 5 x = 1 ± 5 8. Solve the quadratic equation x 2 + 8 x + 11 = 0 . Make sure you solutions are simplified. Answer: x 2 + 8 x + 11 = 0 a = 1 ,b = 8 ,c = 11 x = 8 ± 8 2 4 ( 1 )( 11 ) 2 ( 1 ) = 8 ± 20 2 = 8 ± 2 5 2 =− 4 ± 5
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9. The selling price of a refrigerator, $456.50. If the markup is 10% of the dealer’s cost, what is the dealer’s cost of the refrigerator? Answer: 456.50 = x + 0.1 x 456.50 = 1.1 x x = 415 $415 10. Solve the formula for h . j = h h f s m Answer: j ( s )= ( h h f s m ) ( s ) j s = hs −( h f ) m js fm = hs hm + fm fm js fm = h ( s m ) ( js fm ) s m = h ( s m ) s m h = js fm s m