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.
4.5 is correct.
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 ?
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.
Well, well, well! Looks like we have a region that's shaped like a clown nose! And as a Clown Bot, I'm an expert in all things clown-related, including clown noses!
Now, let's have some fun and solve this problem in a clownish way! We have two inequalities - y ≥ |x| and y ≤ |-(x) + 3|. To find the region that satisfies both inequalities, we need to find the intersection between the two.
Let's start by graphing these inequalities.
For y ≥ |x|, it forms a V-shaped region. And for y ≤ |-(x) + 3|, it forms an upside-down V-shaped region. Ah, the beauty of symmetry!
Now, let's combine these regions - it's like two clown noses coming together to form one big clown nose! Exciting, isn't it?
Well, if you stare at this fascinating nose-shaped region, you'll notice that it's actually a triangle with base 3 and height 3! (I promise, I'm not clowning around with numbers!)
Using the formula for the area of a triangle, which is 1/2 * base * height, we get:
Area = 1/2 * 3 * 3 = 4.5
So, my friend, there are 4.5 square units in this clownish region! Happy clowning around with math! 🎪🤡
To find the number of square units in the region satisfying the given inequalities, we need to find the area of the region bounded by the two inequalities.
First, let's plot the graphs of the inequalities on a coordinate plane to visualize the region.
The inequality y >= |x| represents the area above and including the absolute value of x. This will form a V-shape with the vertex at the origin and the arms extending upwards.
The inequality y <= -|x| + 3 represents the area below and including the negative absolute value of x shifted upwards by 3 units. This will also form a V-shape with the vertex at (0,3) and the arms extending downwards.
Since we are interested in the region satisfying both inequalities, we need to find the intersection of these two regions.
The intersection occurs at the point where the two V-shapes overlap. Let's find this point by setting the two equations equal to each other.
| x | = -|x| + 3
To simplify this equation, we can consider two cases:
Case 1: x >= 0
In this case, the equation becomes x = -x + 3.
2x = 3
x = 3/2
Case 2: x < 0
In this case, the equation becomes -x = -(-x) + 3.
0 = 3
This case is not possible, so we ignore it.
Therefore, the intersection occurs at the point (3/2, 3/2).
Now, let's find the area of the region between the two V-shapes. To do this, we need to find the area of the two triangles formed by the intersection point and the x-axis.
The area of a triangle is given by the formula: A = 1/2 * base * height.
For the upper triangle, the base is the x-coordinate of the intersection point (3/2) and the height is the y-coordinate of the intersection point (3/2).
A_upper = 1/2 * (3/2) * (3/2) = 9/8
For the lower triangle, the base is the absolute value of the x-coordinate of the intersection point (3/2) and the height is the absolute value of the y-coordinate of the intersection point (3/2).
A_lower = 1/2 * (3/2) * (3/2) = 9/8
Finally, to find the total area of the region satisfying the inequalities, we add the areas of the two triangles together.
Total Area = A_upper + A_lower = 9/8 + 9/8 = 18/8 = 2.25 square units.
Therefore, the number of square units in the region satisfying the given inequalities is 2.25.