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Algebra And Trigonometry I

by: Dorothea Bode

Algebra And Trigonometry I MA 15300

Marketplace > Purdue University > Mathematics (M) > MA 15300 > Algebra And Trigonometry I
Dorothea Bode
GPA 3.97


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This 11 page Class Notes was uploaded by Dorothea Bode on Saturday September 19, 2015. The Class Notes belongs to MA 15300 at Purdue University taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/208123/ma-15300-purdue-university in Mathematics (M) at Purdue University.

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Date Created: 09/19/15
MA 153 Lesson 34 Outline Practice problems for distance learning J our Quiz 1 Solve each system using either elimination or substitution Express your answer as an ordered pair If no solution exists write none 2x 3y 9 Multiply the top equation by 3 and the bottom by 2 3x 2y 11 one of many options 6x 9y 27 6x 4y 22 Add the left sides and right sides to eliminate x 5 y 5 y 1 Substitute y into one of the previous equations to nd x 2x319 2x6 x3 3x y8 b 9x 3y6 Multiply the top equation by 3 9x 3y 24 9x 3y 6 Add the left sides and right sides 018 The lines are parallel and therefore no solution exists Answering of homework questions over lesson 33 Lesson 34 Section 51 Inverse Functions Onetoone function Different x inputs always give different y s outputs Use the horizontal line test to determine onetoone function Example Example Not a function Is a function but not a oneto y one function y A Fails the vertical line test one x maps to two yas Fails the horizontal line test Two x s map to the same y Example Is the following function onetoone a fx2x l b fxlxl Inverse Functions Notation f1x Example yfx yr1ltxgt Domain Range Domain Range In order for the inverse to exist f x must be a onetoone function Example Find the inverse function f 71x a fx 2x3 2x1 b fx 3x5 c fx1 5x2 xSO d fxJ MA 153 Lesson 20 Outline Practice problems for distance learning OIA39 AJA Quiz 1 Find the slope of the line perpendicular to the line y x1 2 Slope of glven 11ne 3 Slope of hue perpendlcular 1s negatlve reelprocal 0 E 2 Find the general form ax by C for the equation of a line through A 43 and B24 3 4 1 1 m 4 2 6 y 3lx4 6 6y 36x4 6y 18 x4 x 6y 22 Answering of homework questions over lesson 20 Lesson 20 Section 34 Functions Domain Inputs so outputs are real E Range outputs operation Output F x Example Is a function Each x maps to only one y y Example Not a function One x maps to two y s y Finding Function Values Example Given x2 3 nd a f 2 b fah c fa Example Given gx 4 5x nd a g a 1 W b Example Given x2 x1 nd and simplify fah fa h Finding Domain All inputs so outputs are real 1 denominators 0 2 Jnegative is not real Example Find domain Express your answer in interval notation a fxx2 x1 4x b fx 3x1 c hx xii 94 x2 d gx J Section 31 2 It forms a star 6 A0 4 B4 0 C0 4 D4 0 E2 2 F 2 2 10 a 415 7 b 14 a 241 b 727 22 Show that dA C c103 C 513 Section 32 4 Line xint 15 0 yint 0 3 12 Horizontal parabola xint 4 0 yint0 42 32 It is the upper half of the circle x y2 4 with center 0 0 and r 2 34 It is the left half ofthe circle x2 y2 25 with center 00 and r5 36 x42y7129 46 xlzy74z 20 50 CS 0 r 17 70 Find the distance between the two stations using the Pythagorean theorem and compare that to the sum of their signal strengths Section 33 2 m 5 14 All four lines travel throuin the origin Those lines with positive slopes go up to the right and those lines with negative slopes to up to the left 22 a y 2 b x 4 24 2x 3y 14 32 2x 3y 7 6 17 36 y 7 5 x 5 38 3x 4y 21 58 a P 125t 8250 b t 26 months c The endpoints of the graph are 0 8250 and 66 0 62 a Tut118 99 b During the year 1910 a F 0 b C160andF320 Section 34 4 f 2 f0 0 f3is undefined 6 a 4a3 b 4a3 c 4a73 d 4a74h3 e 4a4h6 f 4 10 a 2a23a77 b 2112731177 c72a273a7 d 2a24ah2h23a3h77 e 2a22h23a3h714 f 4a2h3 12 M b 4 a 21175 c 2J3975 d 21175 20 a 757 b 712 c f1711 d x737135 e 3lU35 28 2 u2 00 b DooooRloo c Decreasing on 00 0 Increasing on 0 00 54 fx7x4 68 3 yOCi b Sx3x42 x x 76 a Lx12500x722 b approx 579 feet 2545 2 ft Section 35 4 Even 6 Odd 10 Neither 14 Given gx l x l and fx 1x cl To find fx For c 3 shift gx left 3 units For c 1 shift gx right 1 unit For c 3 shift gx right 3 units 16 Given gx 2x2 and fx 2x2 c to find fx For c 4 shift gx up 4 units For c 2 shift gx down 2 units For c 4 shift gx down 4 units 32 1 8 38 graph of f horizontally stretched by 2 and shifted down 3 42 Given fx as drawn a shift f right 2 units shift f left 2 units shift f down 2 units shift f up 2 units re ect f through the xaxis and vertically stretch it by a factor of 2 re ect f through the xaxis and vertically compress it by a factor of 2 re ect f through the yaxis and horizontally compress it by a factor of 2 horizontally stretch f by a factor of 2 1 re ect f about the xaxis shift it left 4 units and down 2 units shift f right 4 units and up 2 nap057 239 39 Fr 01x if x g 500000 66 Tx 0125x 71250 ifx gt 500000 00577x if 0 g x g 1000 68 Cx 45000532x 150000511x ifxgt5000 if 1000 lt x S 5000 Section 36 10 fx 4x 22 3 14 a x6 0 b 9 is a maximum c graph is a parabola that opens down with vertex at 3 9 16 a x 7E 3 MIL b 2604 is a minimum c Graph is a parabola with vertex 12 24 opens up and has the xintercepts obtained from part a 24 yx2 4 5 x 26 y 9 12 2 7 2 32 74 77 y 64x 46 125 yd by 250 yd 139 Section37 46 a yfx722 a 4 b 14 by fx c 45 d 72 c 7 x42 5 50 y f 10 a 3x26x3 b 358 1 39 c 274 d x2 18 a 27x318x2 b 3x36x2 c 144 d 135 o gt 52 64 a D62R52 b D 3 1 R 10 4 c D40R 51 d D106 115 e D26R 10 4 f D62 410


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