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UWSpring

MATH136 Midterm: Sample

20 Page
20 Jan 2020
Student name (print legibly) (family name) (given name) Instructions: fill in your waterloo id number and quest login in the box at the top of this pag
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UWSpring

MATH136 Chapter 2: A2_Soln_F17

5 Page
20 Jan 2020
= (5)(7) + ( 2)(6) + ( 3)(8) + (2)( 5) = 11. = p22 + 12 + ( 4)2 = 21. (2) find the scalar equation for the plane with vector equation. To write the sca
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UWSpring

MATH136 Midterm: Sample

22 Page
20 Jan 2020
Student name (print legibly) (family name) (given name) Instructors/sections (cid:3) 001 andrew beltaos (11:30) (cid:3) 002 linda farczadi (1:30) (cid:
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UWSpring

MATH136 Chapter 3: A3_Soln_F17

10 Page
20 Jan 2020
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UWWinter

MATH138 Lecture Notes - Lecture 2: Solution Set, The Roots

6 Page
8 Jan 2020
Thus the solution to |x 2| = 5 is (b) |x 4| = |3x + 2| In this case we have that |x 4| = (x 4) and |3x + 2| = (3x + 2) so that our equation becomes. (x
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UWWinter

MATH138 Lecture 2: Assign1 - 137 - F19

1 Page
8 Jan 2020
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UWSpring

MATH135 Midterm: A01

7 Page
8 Aug 2019
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UWSpring

MATH135 Study Guide - Midterm Guide: Natural Number

7 Page
8 Aug 2019
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UWSpring

MATH135 Study Guide - Midterm Guide: Natural Number

7 Page
8 Aug 2019
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UWSpring

MATH135 Study Guide - Midterm Guide: Order Of Merit

6 Page
8 Aug 2019
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UWMATH104Steve SpencerFall

Lecture1a.pdf

9 Page
7 Oct 2013
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UWMATH104Steve SpencerFall

MATH104 Chapter Notes -Parallelogram Law

3 Page
7 Oct 2013
So far, we have always had our vectors start at the origin, and end at the point corresponding to our vector. But if we are thinking of vectors as a di
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UWMATH104Steve SpencerSummer

MATH104 Chapter Notes -Parametric Equation

2 Page
7 Oct 2013
Vector equation of a line in r2 (pages 5-6) The equation x2 = (0. 5)x1 + 1 de nes the x2 component of a point on the line in terms the x1 component. Bu
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UWMATH104Steve SpencerFall

Lecture1a.pdf

9 Page
7 Oct 2013
View Document
UWMATH104Steve SpencerFall

MATH104 Chapter Notes -Parallelogram Law

3 Page
7 Oct 2013
So far, we have always had our vectors start at the origin, and end at the point corresponding to our vector. But if we are thinking of vectors as a di
View Document
UWMATH104Steve SpencerSummer

MATH104 Chapter Notes -Parametric Equation

2 Page
7 Oct 2013
Vector equation of a line in r2 (pages 5-6) The equation x2 = (0. 5)x1 + 1 de nes the x2 component of a point on the line in terms the x1 component. Bu
View Document

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UWMATH104Steve SpencerFall

Lecture1a.pdf

9 Page
7 Oct 2013
View Document
UWMATH104Steve SpencerSummer

MATH104 Chapter Notes -Parametric Equation

2 Page
7 Oct 2013
Vector equation of a line in r2 (pages 5-6) The equation x2 = (0. 5)x1 + 1 de nes the x2 component of a point on the line in terms the x1 component. Bu
View Document
UWMATH104Steve SpencerFall

MATH104 Chapter Notes -Parallelogram Law

3 Page
7 Oct 2013
So far, we have always had our vectors start at the origin, and end at the point corresponding to our vector. But if we are thinking of vectors as a di
View Document

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