How many square units are in the region satisfying the inequalities y>=(x) and y<=-(x)+3? Express your answer as a decimal. * the () are absolute value signs.

Answer 1

y ≥ x is the region above and including y = x
y ≤ -x + 3 is the region below and including y = -x + 3

I see a region open at the left, so we can't find the area.
We need to close it up on the left.
Are we looking at the little triangle between the two lines and the y-axis ?

Answer 2

5.85

Answer 3

POo fatty lol you are a dummy boi u stupido fatty

Answer 4

4.5 is correct.

Answer 5

The graph of the two inequalities is shown below:

[asy]
Label f;

f.p=fontsize(4);

xaxis(-3,3,Ticks(f, 1.0));

yaxis(-0,4,Ticks(f, 1.0));

fill((0,0)--(-1.5,1.5)--(0,3)--(1.5,1.5)--cycle, grey);
draw((0,0)--(-3,3), Arrow);
draw((0,0)--(3,3), Arrow);
draw((0,3)--(-3,0), Arrow);
draw((0,3)--(3,0), Arrow);
label("$A$", (-1.5,1.5), W);
label("$B$", (0,3), N);
label("$C$", (1.5,1.5), E);
label("$D$", (0,0), S);
[/asy]

The shaded region is the solution set to the two given inequalities. Angle $ADC$ is a right angle because $\overline{AD}$ has slope -1 and $\overline{DC}$ has slope 1, and the two slopes are negative reciprocals. Similarly, the other three angles between sides bounding the shaded region are also right angles. Since $AD=DC$ by symmetry, $ABCD$ is a square. A diagonal of the square is $BD$, which measures 3 units. So a side of the square measures $3/\sqrt{2}$ units and the area is $(3/\sqrt{2})^2=\boxed{4.5}$ square units.