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trasher [3.6K]
3 years ago
14

2:x and 12:18 a-3, b-4, c-6

Mathematics
2 answers:
Bezzdna [24]3 years ago
5 0
2/x = 12/18
cross multiply
(12)(x) = (18)(2)
12x = 36
x = 36/12
x = 3
nirvana33 [79]3 years ago
3 0

The Corrected Problem is :

2:x and 12:18 identify the value of x that makes each pair of ratios equivalent .

Solution:

If a pair of Ratios are equivalent then we can write

\frac{2}{x}=\frac{12}{18}\\ \\ \text{Simplify we get}\\ \\ 12x=36\\ \\ \text{Divide both the sides by 12 we get}\ \\ \\ \frac{12x}{12}=\frac{36}{12}\\ \\  x=3\\ \\ \text{Hence the required value of x is 3}\\

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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.

7 0
3 years ago
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