Solve the following inequality giving brainliest
1 answer:
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<span>We have a rectangle with two sides a and b, with length three times its width. So let's name:
a = </span>length
b = width
So, according to the relation:
Part a)
First of all we draw the rectangle as shown in the figure 1. To place this rectangle on the coordinate plane we will take a point (x, y) and name it v1 that is the vertex 1. Next we will name the other vertices as v2, v3 adn v4. Finally, the segment v1v2 must be parallel to the x-axis
Part b)
Given that the vertex v1(x, y). As shown in the figure 2, v2 has the same value of y, but x increases 3b, so v2(x+3b, y). Applying the same reasoning to v3 and v4, we hava:
v1(x, y)
v2(x+3b, y)
v3(x, y+b)
v4(x+3b, y+b)
a = </span>length
b = width
So, according to the relation:
Part a)
First of all we draw the rectangle as shown in the figure 1. To place this rectangle on the coordinate plane we will take a point (x, y) and name it v1 that is the vertex 1. Next we will name the other vertices as v2, v3 adn v4. Finally, the segment v1v2 must be parallel to the x-axis
Part b)
Given that the vertex v1(x, y). As shown in the figure 2, v2 has the same value of y, but x increases 3b, so v2(x+3b, y). Applying the same reasoning to v3 and v4, we hava:
v1(x, y)
v2(x+3b, y)
v3(x, y+b)
v4(x+3b, y+b)