Find the coordinates of the vertices of the figure formed by each system of inequalities. x + y < 9 x - 2y < 12 y < 2x + 3 a. (0, 0), (0, 1), (2, 0) c. (9, 1), (2, 3), (5, 6) b. (10, -1), (-6, -9), (2, 7) d. (0, -1), (2, -1), (3, 6)

b. (10, -1), (-6, -9), (2, 7)

Step-by-step explanation:

One way to solve this problem is by graphing (see attachment). In fact, that actually is the easiest way to do this.

First, graph x + y < 9 (red part on the attachment). To do this, isolate the y so that it's like a regular linear equation: y < -x + 9. Graph the line y = -x + 9 first, and then choose a point either on the left or right of the line to plug in for x and y.

Let's just choose (0, 0), which is on the left of the line: 0 < -0 + 9  ⇒  0 < 9. Since this is a true statement, we know that we should shade the left side of the equation, and not the right side.

Now, graph x - 2y < 12 (blue part on the attachment). It's the same process here as the first inequality. For this one, we find that we also shade to the "left" of the line (it's more "up" than "left", but you get the idea).

Finally, graph y < 2x + 3 (green part on the attachment). Again, it's the same process. For this, we shade to the right.

Now look at the three points that they intersect at: they are (2, 7), (-6, -9), and (10, -1). The answer is thus B.

Step-by-step explanation:

Step 1:  Graph each of the inequalities

x + y < 9

x - 2y < 12

y < 2x + 3

It comes out that the vertices are (2, 7), (10, -1), and (-6, -9)

 Option B

14

step-by-step explanation:

14 x 3 = 42

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.