Move all terms not containing
|
5
−
8
x
|
|
5
-
8
x
|
to the right side of the inequality.
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Add
7
7
to both sides of the inequality.
|
5
−
8
x
|
<
8
+
7
|
5
-
8
x
|
<
8
+
7
Add
8
8
and
7
7
.
|
5
−
8
x
|
<
15
|
5
-
8
x
|
<
15
Remove the absolute value term. This creates a
±
±
on the right side of the inequality because
|
x
|
=
±
x
|
x
|
=
±
x
.
5
−
8
x
<
±
15
5
-
8
x
<
±
15
Set up the positive portion of the
±
±
solution.
5
−
8
x
<
15
5
-
8
x
<
15
Solve the first inequality for
x
x
.
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x
>
−
5
4
x
>
-
5
4
Set up the negative portion of the
±
±
solution. When solving the negative portion of an inequality, flip the direction of the inequality sign.
5
−
8
x
>
−
15
5
-
8
x
>
-
15
Solve the second inequality for
x
x
.
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x
<
5
2
x
<
5
2
Set up the intersection.
x
>
−
5
4
x
>
-
5
4
and
x
<
5
2
x
<
5
2
Find the intersection between the sets.
−
5
4
<
x
<
5
2
-
5
4
<
x
<
5
2
The result can be shown in multiple forms.
Inequality Form:
−
5
4
<
x
<
5
2
-
5
4
<
x
<
5
2
Interval Notation:
(
−
5
4
,
5
2
)
(
-
5
4
,
5
2
)
Answer:
when the square on the hypotenuse of a right-angled triangle is equal in to the sum of the squares on the other two sides
Step-by-step explanation:
let's say......
h = ?
o = 4
a = 4
a^2+o^2=32
square root 32
5.65685424949
that is the value for h
Find the distance between the points to the nearest tenth.
1. C(-1, -1), D(6, 2)
<h3>= 7.6</h3>
2. E(-7, 0), F(5 ,8)
=3(3.6)
<h3>=10.8</h3>
______________________________________________________
AB has endpoints A(-3, 2) and B(3, -2).
3. find the coordinates of the midpoint of Line AB
4. Find AB to the nearest tenth
<h3>
AB = 7.2 units.</h3>
I’m not really sure if it correct, g(1)=0 or g(1)=1??