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:
The equation is equal to each other it ends up simplifying to
8a-39=8a-39
they are both equal to each other but if you simplify it even more thye undo each other so it becomes
0=0
Step-by-step explanation:
Solution:
we are given that
Nathan takes 42 minutes to thread a distance of 4.5 miles on a threadmill.
we have been asked to find the average distance covered in one minute.
Using the concept of unity we can write
Since Nathan takes 42 minutes to cover a distance of 4.5 miles.
So in 1 minute Nathan will cover a distance of 
Hence avearge distance covered in one minute is 1.1 miles.
Ration or ratio? I'm assuming ratio. 12:13 I think.... let me know if I'm wrong
Answer:
Solution to determine whether each of these sets is countable or uncountable
Step-by-step explanation:
If A is countable then there exists an injective mapping f : A → Z+ which, for any S ⊆ A gives an injective mapping g : S → Z+ thereby establishing that S is countable. The contrapositive of this is: if a set is not countable then any superset is not countable.
(a) The rational numbers are countable (done in class) and this is a subset of the rational. Hence this set is also countable.
(b) this set is not countable. For contradiction suppose the elements of this set in (0,1) are enumerable. As in the diagonalization argument done in class we construct a number, r, in (0,1) whose decimal representation has as its i th digit (after the decimal) a digit different from the i th digit (after the decimal) of the i th number in the enumeration. Note that r can be constructed so that it does not have a 0 in its representation. Further, by construction r is different from all the other numbers in the enumeration thus yielding a contradiction
