Which statements are true about the linear inequality y > 3/4x – 2? Check all that apply. The slope of the line is –2. The gr
aph of y >3/4 x – 2 is a dashed line. The area below the line is shaded. One solution to the inequality is (0, 0). The graph intercepts the y-axis at (0, –2).
1 answer:
1. Consider the function 
.
It is a linear function, so to draw its graph 2 points are enough:
f(4)=(3/4)*4-2=3-2=1
f(0)=(3-4)*0-2=-2
so 2 points on this line are (0, -2) and (4, 1). Check the picture 1:
2. The problem considers the inequality y>f(x)
a. the slope of the line y=ax+b is the coefficient of x, so a. In our case a=3/4, not -2
b. The graph of any linear inequality is a shaded region, not a line or a shaded line.
c. Either the part below the line, or the part above the line will be shaded since we are talking about an inequality.
Check (4, -3), picture 2, which is clearly below the line.
now here y=-3, and f(4)=(3/4)(4)-2=3-2=1, so in this region y<f(x)=(3/4)x-2
So clearly the region where the inequality holds, so the region to color, is the part above the line.
d. The line itself is not dashed, only if we have an "equal to or smaller than" or "equal to or larger than" inequality.
e. (0, 0) is in the region above the line so YES, it is a solution.
We can also check it like this: y=0, f(0)=(3/4)0-2=-2 so y>f(0)
f. The y-intercept is always f(0). In this case f(0)=-2
It is a linear function, so to draw its graph 2 points are enough:
f(4)=(3/4)*4-2=3-2=1
f(0)=(3-4)*0-2=-2
so 2 points on this line are (0, -2) and (4, 1). Check the picture 1:
2. The problem considers the inequality y>f(x)
a. the slope of the line y=ax+b is the coefficient of x, so a. In our case a=3/4, not -2
b. The graph of any linear inequality is a shaded region, not a line or a shaded line.
c. Either the part below the line, or the part above the line will be shaded since we are talking about an inequality.
Check (4, -3), picture 2, which is clearly below the line.
now here y=-3, and f(4)=(3/4)(4)-2=3-2=1, so in this region y<f(x)=(3/4)x-2
So clearly the region where the inequality holds, so the region to color, is the part above the line.
d. The line itself is not dashed, only if we have an "equal to or smaller than" or "equal to or larger than" inequality.
e. (0, 0) is in the region above the line so YES, it is a solution.
We can also check it like this: y=0, f(0)=(3/4)0-2=-2 so y>f(0)
f. The y-intercept is always f(0). In this case f(0)=-2
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Follow along and see the highlighted bold text for the drop-down answers.
Our first option is asking for the expression which is the sum of the area of:
a. the shaded triangles
and b. the white square
The area of the shaded triangles can be found by using the formula for the area of a triangle.
A = (base * height) / 2
On these triangles, the base is b and the height is a.
Since there are 4 triangles, we'll multiply this expression by 4.
The area of the white square can be found using the formula for the area of a square.
A = side * side = side^2
The side of our square is c.
The sum of these areas is, thus:
(our first expression)
The area of the larger square, the entire figure, can be found using the formula for the area of a square that we used above. The side of this square is (a + b):
(our 2nd expression)
We also choose to expand the parenthesis using the appropriate rule to get the above expression.
Now, following the proof, we set both of these expressions equal to each other and subtract something. Let's put these 2 expressions (the sum of the triangle and the white square, and the larger square) equal to each other:
Since 2ba = 2ab, we can subtract this from both sides. (we subtract 2ab)
Flipping this equation, we get the Pythagorean Theorem: