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daser333 [38]
3 years ago
5

Jim says that the output of the floor function is the number before the decimal point in the input. For what domain is Jim’s sta

tement correct?
Mathematics
1 answer:
Ann [662]3 years ago
5 0

Answer:

Jim's statement is correct for positive numbers


Explanation:

Flooring means that we will take the number before the decimal point. This number should be the whole number that precedes the given decimal.


Now, let's check:

1- positive numbers:

Take 3.5 for example:

Taking the number directly before the point ......> we would have 3

The whole number that directly precedes 3.5 is .......> 3

Since they are both the same, therefore, Jim's statement is true


2- negative numbers:

Take -3.5 for example:

Taking the number directly before the point ......> we would have -3

The whole number that directly precedes 3.5 is .......> -4

Since they are both not the same, therefore, Jim's statement is false


Hope this helps :)



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Answer:

See the attached image for the graph of the first system.

Step-by-step explanation:

Here's how to graph the first system.

Start with the inequality -y \le -2x-3.  You can make this easier to work with by multiplying through by -1.  Remember to switch the inequality sign when multiplying by a <u>negative</u> number.  OK, you get the inequality

y \ge 2x+3.

The graph will be a half-plane -- all the points on one side of a line.  The line that is the boundary of the half-plane has an equation:  y=2x+3 -- just use an  =  sign instead of the inequality sign.

Graph the line.

The equation of the line is in slope-intercept form:  y = mx + b, so you can tell the y-intercept is 3 and the slope is 2 (think of it as a fraction 2/1).  Graph the line by going to the point (0, 3) -- the y-intercept -- then use the slope 2/1 interpreted as "rise over run" to go up 2 units and right 1 unit, arriving at the point (1, 5).  Draw the line through those points, (0, 3) and (1, 5).

Now you have to decide which side of the line the inequality y \ge 2x+3 is describing. To do this, pick a point which is not on the line, plug its coordinates into the inequality; if the result is true, shade the side of the line the point you picked is on (if false, shade the <u>other</u> side!)

An easy point to pick in this case is the origin, (0, 0).  Put zeros in for x and y in the inequality, and you'll get the statement 0 \ge 2(0)+3 \, \Rightarrow \, 0 \ge 3.  That's <u>false</u>, so shade the side of the line <u>not</u> containing the origin.  In the image below, the shading is in purple.

All right, now for the other inequality, x+2\le 0.  Subtract 2 from both sides and the inequality becomes x \le -2.  This, too, graphs as a half-plane whose boundary line has equation x=-2.  Graph the line.  A line with an equation that has  x  in it but not  y is a vertical line with all its x-coordinates equal to the number on the right side of the equation.  This line is vertical and goes through points such as (-2, 0).

Pick a point <u>not</u> on the line (the origin works again).  Put the coordinates into the inequality to get 0\le -2 which is <u>false</u>.  Shade the side of the vertical line which does <u>not</u> contain the origin.  In the image below, the shading is in black.

Finally,  YAY!  \o/ ,  the solutions to the system are all the points in the plane that got shaded twice.  In the image, they are the cross-hatched points above the purple line and to left of the black line.

Note: If you get a system with three inequalities, you'll be graphing three half-planes and looking for points that got shaded three times!

Note: One of your questions has the inequalities x \ge 0 and y \ge 0 in it.  These two inequalities say that the x and y coordinates are both positive or zero, confining your attention to Quadrant I in the upper-right part of a graph, above the x-axis <u>and</u> to the right of the y-axis.

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