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step 1 first u should seclect the name independent and dependent veriables
2. take X axis ur independent veriable and Yaxis as a dependent veriable
3.lable the axis with units
4. now draw all points on the paper
5.connect first and last point of the reading
6. make sure that u have slected a suitabe scale for drawing the graph of a complecated data
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When graphing an inequality with one variablehow do you graph less than or greater than problem?
Arrange the inequality so that the variable is on the left. ex x < 7 If not equal to put an open circle at the number (7 in my example) if less than shade the number line to the left ( less than = shade left) if greater than shade right. If equal to put a point ( shaded dot) on the number follow same rules for shading
Is -4 a polynomial?
is -4 a polynomial? This depends on what you accept as a definition A polynomial is often defined as a set of things in order obeying certain rules. ( these things and rules can be very complicated) A polynomial EQUATION is an equation between two polynomials When using only real numbers and "regular" math rules -4 is a polymomial of degree 0 x = -4 is a polynomial equation is a polynomial of degree 1 it is the same as x +4 = 0 It can be represented by { 4, 0} Sometimes the terms are used interchangably
What Allows you to divide both sides of a equation by the same number?
The rules of algebra: more specifically, it is the the existence of a multiplicative inverse for all non-zero values.
Does raising each side of a compound inequality to a negative exponent flip the signs of the inequality?
Yes, taking the reciprocal (raising each side to the -1 power) of each side of a compound inequality can flip the signs of the inequality. This can be useful when you have an inequality with 'x' in the quotient. Taking the reciprocal of each side can be a more direct way of solving the inequality than multiplying each side by 'x'. The following is an example: | 2/x - 2 | < 4 Following the rules for an absolute value inequality we obtain the following compound inequality: -4 < 2/x - 2 < 4 Next add 2 to each side to get 'x' by itself. -2 < 2/x < 6 Here we can multiply each side by 'x' to deal with 'x' in the quotient, but instead we'll raise each side to an exponent of (-1). We obtain the following: -1/2 > x/2 > 1/6 (Notice the signs flip.) We rewrite as: 1/6 < x/2 < -1/2 Next multiply each side by 2 to get 'x' by itself. 1/3 < x < -1 Our solution set is the following: {x: x > 1/3 OR x < -1} Which is the union of the two infinite intervals (-infinity, -1) AND (1/3, infinity). For these types of inequalities if we believe that perhaps we've made a mistake or that our signs are wrong, we can check our work by plugging in some values for x and evaluating the inequality to see whether or not the statement is true. It helps to graph the inequality on a line and by evaluating x at different points on the graph of our inequality for the values of x that make our statement true; we can see exactly what the inequality looks like. For example, we will evaluate the original inequality with points that are less than -1, in between -1 and 1/3, and greater than 1/3. We'll try x = -2 first, |2/(-2) - 2| < 4 |-1-2| < 4 |-3| < 4 -(-3) < 4 3 < 4 True, our solution: x < -1 holds true. Next we'll solve for x = -1/2, |2/(-1/2) - 2| < 4 |-4 - 2| < 4 |-6| < 4 -(-6) < 4 6 < 4 False, this point is not on the graph of our inequality, so we know that the sign of our solution: x < -1 is going in the right direction and holds true. Next we'll solve for x = 1/4, |2/(1/4) -2| < 4 |8 - 2| < 4 |6| < 4 6 < 4 False, this point is not on the graph of our inequality, so it looks like our second solution x > 1/3 is accurate and our sign is most likely going in the correct direction. Lastly, we'll evaluate for a point x > 1/3 and this point should be on the graph of our inequality. |2/(1) - 2| < 4 |2-2| < 4 |0| < 4 0 < 4 True, we've proved that our solution x > 1/3 holds true for the graph of this inequality and that the sign for our solution is going in the correct direction. In fact if we substitute a very large number in for x, say 1,000 we'll notice the left side of our statement gets closer and closer to 2 as x approaches infinity. |2/(1000) -2| < 4 |-1.998| < 4 -(-1.998) < 4 1.998 < 4 True, we know for certain that the solution x > 1/3 holds true for all values of x to infinity. Our solution set again is, {x: x < -1 or x > 1/3} The union of the two infinite intervals is (-infinity, -1) and (1/3, infinity).
How are recursive rules different from explicit function rules for modeling linear data?
recursive rules need the perivius term explicit dont
