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by: Alex Tucker

20

1

6

# MATH 95 - Week 1 Notes MATH 95

Marketplace > University of Oregon > Math > MATH 95 > MATH 95 Week 1 Notes
Alex Tucker
UO

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This is what we covered during Week 1.
COURSE
Math 95
PROF.
Craig Tingey
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Lines, Substitution
KARMA
Free

## Popular in Math

This 6 page Class Notes was uploaded by Alex Tucker on Saturday October 1, 2016. The Class Notes belongs to MATH 95 at University of Oregon taught by Craig Tingey in Fall 2016. Since its upload, it has received 20 views. For similar materials see Math 95 in Math at University of Oregon.

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Date Created: 10/01/16
Lines (Day 2) 9/27/16 lin Parallel e lines intersectio (Parallel Lines never n cross) Y axis Perpendicular lines (horizontal, vertical) (-2,0) (0,1) X axis The points on a line satisfy the equation of the form Ax + By + C = 0. rise change(y) Slope= Slope = m = (y 1y 2/ run change(x) (x1-x2) Lines (Day 2) 9/27/16 M = -1 M = 1 M = 2 M = ½ (bigger slopes = Equations: - Standard Form Ax + By + C = 0 - Slope-Intercept y=mx + B - Slope = m - Y-int = b (0,b) - Point-Slope y-y = m (x-x ) 1 1 - Slope = m - (x1,1 ) = any point Example: Find an equation of the line through (-1,2) and (3,1). 2−1 −1 −1−3 = 4 y-2 = -¼(x-(-1)) + - slope slope Lines (Day 2) 9/27/16 Perpendicular lines have slopes which are negative reciprocals of each other. 1 m⊥= m Applications of Linear Functions (Day 3) 9/28/16 Doritos cost \$3.29 / bag. How much does it cost to buy x bags? Call the cost C=C(x) C(x)=3.29x+0 More generally, total cost = (fixed costs) + (variable costs) Ex. Shirts (flat fee) \$50 for stencil + \$15 per shirt; total cost for x shirts: C(x)=15x + 50 - Systems of Linear Equations Let’s find out where the lines x+ 5y = 3 & 3x + 4y = -2 intersect. x+5y=3 3x+4y=-2 -x -x -3x -3x 5y −x+3 4y=−3x−2 = 5 5 4 −1 3 −3 1 y= 5 x+ 5 y= 4 x− 2 Applications of Linear Functions (Day 3) 9/28/16 - Substitution 1) Solve one of the equations for one variable. 2) Substitute into other sign. 3) Use formula from one to find remaining variable. I’ll solve x+5y=3 for x. x= -5y+3  3(-5y+3)+4y= -2 x=-5+3; x = -2 -15y+9+4y= -2  -11y+9=-2 -9 -9 −11 y=−11 11 y = 1 (-2,1) A. Solve 2x+y=1 & 3x+2y=0 y=-2x+1 3x+2(-2x+1)=0 3x-4x+2=0  -x+2=0 −x=−2 −1 x = 2 Applications of Linear Functions (Day 3) 9/28/16 2(2)+y=1  4+y=1 -4 -4 y=-3 (2,-3) B. Solve 3x+2y=4 & 6x+4y=-1 2y=−3x+4 2 y=−3 x+2 2 −3 6 x+4(2 x+2)=−1 6x-6x+8=-1  8 = -1? No Solutions

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