Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
3. Multiplying Binomials
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Exercise 54 Page 503

Practice makes perfect
a We are given the system of inequalities which determines the trapezoid.

y≥ 3 & (I) y≤ 9 & (II) x≤ 8 & (III) y≤ 2x+3 & (IV) We will graph the first two inequalities together, and the other two separately. Next, we will combine the graphs. The trapezoid will be the overlapping region.

Inequalities I and II

We can write the equation of the boundary line by changing the inequality symbol to an equals sign. Let's do it for Inequality I and Inequality II. Inequality I:&y≥ 3 Boundary Line I:&y=3 Inequality II:&y≤ 9 Boundary Line II:&y=9 Since both inequalities are not strict, both boundary lines are solid. What is more, both boundary lines are horizontal.

Our inequalities are y≥ 3 and y≤ 9. This means that every coordinate pair with y-value greater than or equal to 3 and less than or equal to 9 needs to be included in the shaded region.

Inequality III

The boundary line for the third inequality is x=8. This is a vertical line. The inequality is not strict, so the boundary line is solid.

The inequality x≤ 8 describes all values of x that are less than or equal to 8. This means that every coordinate pair with an x-value that is less than or equal to 8 needs to be included in the shaded region.

Inequality IV

Let's write the equation of the boundary line. Inequality:&y≤ 2x+3 Boundary Line:&y=2x+3 Since this equation is in slope-intercept form, we can determine its slope m and y-intercept b to draw the line. Slope-Intercept Form:&y=mx+b Boundary Line:&y=2x+3 Now that we know the slope and y-intercept, let's use these to draw the boundary line. Notice that once again the inequality is not strict.

To complete the graph, we will test a point that is not on the line and decide which region we should shade. Let's test the point ( 0, 0). If the point satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
y≤ 2x+3
0? ≤ 2( 0)+3
â–Ľ
Simplify
0? ≤0+3
0≤ 3
Since the point satisfies the inequality, we will shade the region that contains this point.

Combining the Inequality Graphs

Let's draw the graphs of the inequalities on the same coordinate plane.

The overlapping region is the trapezoid.

b We are given the formula for the area A of a trapezoid.
A=1/2(b_1+b_2)h Here b_1 and b_2 are the bases of the trapezoid, and h is the height. To determine the area of the trapezoid graphed in Part A, we have to determine the lengths of its bases, b_1 and b_2, and its height, h.
Looking at our graph, we can see that b_1=8, b_2=5, and h=6. Let's substitute these values into the formula. A=1/2(8+5)6 Finally, we will simplify the expression above.
A=1/2(8+5)6
â–Ľ
Simplify
A=1/2(13)6
A=(13)6/2
A=(13)3
A=39
The area of the trapezoid is 39 square units.