Gr. 10 Math Exam Notes

Topics: Quadratic equation, Triangle, Linear equation Pages: 8 (2712 words) Published: May 29, 2013
Math Exam Notes
Unit 1
The Method of Substitution
-Solving a linear system by substituting for one variable from one equation into the other equation -To solve a linear system by substitution:
Step 1: Solve one of the equations for one variable in terms of the other variable
Step 2: Substitute the expression from step 1 into the other equation and solve for the remaining variable
Step 3: Substitute back into one of the original equations to find the value of the other variable
Step 4: Check your solution by substituting into both original equations, or into the statements of a word problem -When given a question in words, begin by defining how variables are assigned Investigate Equivalent Linear Relations and Equivalent Linear Systems -Equivalent linear equations: equations that have the same graph -Equivalent linear systems: pairs of linear equations that have the same point of intersection -For any linear equation, an equivalent linear equation can be written by multiplying the equation by any real -Equivalent linear systems have the same solution; the graphs of linear relations in the system have the same point of intersection -Equivalent linear systems can be written by writing equivalent linear equations for either or both of the equations, or by adding or subtracting the original equations The Method of Elimination

-Solving a linear system by adding or subtracting to eliminate one of the variables -To solve a linear system by elimination:
-Arrange the two equations so that like terms are aligned
-Choose the variable you wish to eliminate
-If necessary. Multiply one or both equations by a value so that they have the same or opposite coefficient in front of the variable you want to eliminate
-Add or subtract to eliminate one variable
-Solve for the remaining variable
-Substitute into one of the original equations to find the value of the other variable Solving Problems Using Linear System
-You can solve linear systems using any of the three methods: graphing, substitution, or elimination **Solve problems with Distance, Speed, Time**
-Downstream d = (p+c)t
-Upstream d = (p-c)t
**Solve a mixture problem**
-Solution: x + y = amount of mixture
-Pure acid: %x + %y = %mixture
Unit 2
Midpoint of a Line Segment
-Midpoint: a point that divides a line segment into two equal line segments (x1+x2 ÷ 2 = x) (y1+y2 ÷2 = y) -Median: line segment joining a vertex of a triangle to the midpoint of the opposite side -To find an equation for the median of a triangle, first find the coordinates of the midpoint of the side opposite to the vertex. Use the coordinates of the midpoint and the vertex to calculate the slope of the median. Then, substitute the slope and the coordinates of either point into y = mx + b to solve for the medians y-intercept -Equidistant: equally distant

-Right bisector: the line that passes through the midpoint of a line segment and intersects it at a 90° angle -To find an equation for the right bisector of a line segment, first find the slope and midpoint of the segment. Use the line segment’s slope to calculate the slope of a perpendicular line. Then, substitute this slope and the coordinates of the midpoint into y = mx + b to solve for the right bisector’s y-intercept Length of A Line Segment

-You can calculate the length, d, of a line segment using its rise and run: d = √(run)2 + (rise)2 or d = √(x2-x1)2 + (y2-y1)2
Equation for a Circle
-An equation for the circle with centre at the origin and radius r is x2 + y2 = r2 -The radius of a circle centered at the origin is r = √ x2 + y2 Unit 3
Investigate Properties of Triangles
-The medians of a triangle meet at a single pint: centroid
-Each median bisects the area of the triangle
-The median from the vertex, between the equal sides of an isosceles triangle coincide with the altitude to the vertex and bisects the angle at the vertex Verify Properties of Triangles
-The centroid of a triangle divides each median into...
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