Answer:
Both expressions should be evaluated with two different values. If for each substituted value, the final values of the expressions are the same, then the two expressions must be equivalent.
Step-by-step explanation:
Both expressions are linear expressions. It takes 2 points to define a line. If the lines defined by each expression go through the same two points, then the expressions are equivalent.
If the expressions have the same value for two different variable values, they are equivalent. (choice D)
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<em>Additional comment</em>
One more point is needed than the degree of the polynomial expression. That is, quadratic (degree 2) expressions will be equivalent if they go through the same 2+1 = 3 points.
<u><em>The additive identity property </em></u>. . . is being used here because no matter what the value is, whenever you add zero (0) to it, you'll always get the value your started with.
Answer:
A. m = -2
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
- Slope Formula:
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Find points from graph.</em>
Point (0, 0)
Point (1, -2)
<u>Step 2: Find slope </u><em><u>m</u></em>
Simply plug in the 2 coordinates into the slope formula to find slope<em> m</em>
- Substitute in points [SF]:
- [Fraction] Subtract:
- [Fraction] Divide:
The frist line may be one with the numbers 0, 1 and 2 and 10 divisions (marks at equal distance) between each integer. Each division will be equal to 0.1 units and then you can mark the second division from the zero point to the right as the 0.20 mark.
The other line must have the same integers, 0 , 1 and 2 placed in identical form as the first line. Then
- draw an inclined straight line since the point zero,
- mark 5 points in the inclined lined equally spaced over the line.
- draw a sttraight line from the 5th point to the point with the mar 1 over the base number line.
- draw a parallel line to the previous one passing trhough the second point of the inclined line and mark the point where this parallel touchs the base number line. This point shall be at the same distance from zero than the 0.2 mark was in the first number line, meaning that 0.2 and 1/5 are equivalent.
