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Test 12/1015
| Term | Definition |
|---|---|
| BREAK EVEN | COST OF PRODUCTION = THE INCOME |
| COLUMN | NUMBERS ALIGNED VERTICALLY IN A MATRIX |
| CONSISTENT | SYSTEM THAT HAS AT LEAST 1 SOLUTION |
| DEPENDENT | SYSTEM THAT HAS INFINITELY MANY SOLUTIONS |
| DIMENSIONS OF A MATRIX | NUMBER OF ROWS BY THE # OF COLUMNS |
| ELEMENT | EACH OF THE #'S IN A MATRIX |
| ELIMINATION METHOD OR ADDITION METHOD | EQUATIONS ARE ADDED TO ELIMINATE A VARIABLE |
| INCONSISTENT | SYSTEM WITH NO SOLUTION |
| INTERSECTION OF GRAPHS METHOD FOR SOLVING AN EQUATION | GRAPH Y1=LEFT SIDE Y2=RIGHT SIDE. THE X COMPONENT OF THE INTERSECTION IS THE SOLUTION |
| INTERVAL NOTATION | SOLUTION WRITTEN IN THE FORM OF [(2, INFINITY SYMBOL) OR (-INFINITY SYMBOL, 6)] AS EXAMPLES |
| LINEAR INEQUALITY | SIMILAR TO A LINEAR EQUATION EXCEPT THE EQUALITY SYMBOL IS NOW <, >, AND ETC |
| MATRIX | RECTANGULAR ARRAY OF #'S |
| MATRIX ADDITION | WHEN TWO OR MATRICES WITH THE SAME DIMENSIONS ARE ADDED BY ADDING THE ELEMENTS IN THE SAME POSITION TOGETHER |
| ROW | NUMBERS ALIGNED HORIZONTALLY IN A MATRIX |
| SCALAR MULTIPLICATION | WHEN A # IN FRONT OF THE MATRIX IS MULTIPLIED BY EVERY ELEMENT INSIDE THE MATRIX TO GET A PRODUCT WITH THE SAME DIMENSIONS AS THE ORIGINAL MATRIX |
| SET NOTATION | SOLUTION WRITTEN IN THE FORM OF {X|X>2} AS AN EXAMPLE |
| SOLUTION OF A SYSTEM | IN TWO VARIABLES IT IS IN THE FORM OF (X,Y,) IN THREE VARIABLES IT LOOKS LIKE (X,Y,Z). VARIOUS VARIABLES CAN BE USED TO REPRESENT THE SOLUTION. WHEN IT IS GRAPHED IT IS THE INTERSECTION OF THE LINES AT ONE POINT. |
| SOLUTION TO A SYSTEM OF INEQUALITIES | REPRESENTED BY A GRAPH AND SHADING OF ALL POINTS THAT SATISFY ALL LINEAR INEQUALITIES. COMMONLY SHADED AREA ON THE GRAPH. |
| SUBSTITUTION METHOD | TAKE 1 EQUATION AND SOLVE FOR ONE VARIABLE. THE EXPRESSION FOR THAT VARIABLE IS REPLACED FOR THE VARIABLE IN ANOTHER EQUATION |
| SYSTEMS OF EQUATIONS | TWO OR MORE EQUATIONS IN TWO OR MORE VARIABLES ARE SOLVED SIMULTANEOUSLY |
| X-INTERCEPT METHOD FOR SOLVING AN EQUATION | WRITE SO THAT ONE SIDE OF THE EQUATION = O. GRAPH Y1=THE NON-ZERO SIDE OF THE EQUATION. FIND THE X-INTERCEPT OR ZERO OF THE GRAPH. THE X-COORDINATE OF THE X-INTERCEPT IS THE SOLUTION. |
Created by:
ahmari66
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