1. Name: 2. Solve the linear inequality. Express the solution using interval notation and graph the solution set. -5 < 5-2x <-4 13 ,7 Graph: d. Interval: Graph: a. Interval: 13 11 2 Graph: ,5 Graph: Interval: b. Interval: e. 17 6. 2 15 Interval: Graph: c. 8. 15 3. A rectangle has the vertices of A(5, 5), B(8, 5), C(5,-1), and D(8, -1) on a coordinate plane. Find the rectangle. are. 17 a. b. 14 c. 19 d. 22 e. 18 Name: 4. Find the

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter5: Linear Inequalities
Section5.5: Solving Inequalities Involving Absolute Value
Problem 50PFA
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question 3 please

1.
Name:
2. Solve the linear inequality. Express the solution using interval notation and graph the solution set.
-5 < 5-2x <-4
13
,7 Graph:
d. Interval:
Graph:
a. Interval:
13
11
2
Graph:
,5 Graph:
Interval:
b. Interval:
e.
17
6.
2
15
Interval:
Graph:
c.
8.
15
3. A rectangle has the vertices of A(5, 5), B(8, 5), C(5,-1), and D(8, -1) on a coordinate plane. Find
the rectangle.
are.
17
a.
b. 14
c. 19
d. 22
e. 18
Name:
4. Find the
Transcribed Image Text:1. Name: 2. Solve the linear inequality. Express the solution using interval notation and graph the solution set. -5 < 5-2x <-4 13 ,7 Graph: d. Interval: Graph: a. Interval: 13 11 2 Graph: ,5 Graph: Interval: b. Interval: e. 17 6. 2 15 Interval: Graph: c. 8. 15 3. A rectangle has the vertices of A(5, 5), B(8, 5), C(5,-1), and D(8, -1) on a coordinate plane. Find the rectangle. are. 17 a. b. 14 c. 19 d. 22 e. 18 Name: 4. Find the
Expert Solution
Step 1

Given:

A rectangle has the vertices of A(5,5), B(8,5), C(5,-1), and D(8,-1) on the coordinate plane.

104
A(5, 5)
B(8,5)
10
D(8,-1)
15
C(5,-1)
-5+
Step 2

So, from the above graph the area of rectangle is:

.1]
Area = Area of AABD+ Area of AACD
%3D
Step 3

First, we find the area of triangle ABD:

Area of AABD =x, (V, – y; )+ x, ('; – Y;)+ x; (v; – '; ) |
Here, (x; ); ) = (5,5),. (x, V2 ) = (8,5) and (x3. V; ) = (8.–1)
%3D
So,
$(5-(-1) +8(-1-5)+8(5– 5)
Area of AABD =
Area of AABD =5(5+1)+8·(-6)+8·0
30 – 48+ 8.0]
Area of AABD =
x-18
Area of AABD =
Area of AABD =|-9|
|Area of AABD =9
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