Alicia has 32 stickers this is 4 times as many stickers as Benita has how many stickers does Benita have

PLEASE HELP ME ILL GIVE BRAINLIEST SELECT ALL THAT APLY!!!!!!!!!!!!!!!!!!!!!!!

fenix001 [56]

$y\cdot \:2\frac{1}{2} =$ $2\frac{1}{2}=\frac{2\cdot 2+1}{2}=\frac{5}{2}$ $=\frac{5y}{2}$

a) $2y+\frac{1}{2}=\frac{4y+1}{2}$

b) $2y+\frac{1}{2}y=\frac{5y}{2}$

<h3><em>So b) is the expression that is equivallent.</em></h3><h3><em></em></h3><h3><em></em></h3><h3><em>Hope I helped you!</em></h3><h3><em>Success!</em></h3>

8 0

2 years ago

Please help ASAP IM IN DESPREATE NEED OF HELP

Advocard [28]

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.