Sure, let's solve the given problem step-by-step.
Problem Statement:
If ' [tex]\( r \)[/tex] ' is a real number such that [tex]\( |r| < 1 \)[/tex] and if [tex]\( a = 5(1-r) \)[/tex], then:
(A) [tex]\( -5 < a < 5 \)[/tex]
(B) [tex]\( 0 < a < 5 \)[/tex]
(C) [tex]\( 0 < a < 10 \)[/tex]
(D) [tex]\( 0 \leq a < 10 \)[/tex]
To determine the range of [tex]\( a \)[/tex], we start with the given inequality for [tex]\( r \)[/tex]:
[tex]\[ |r| < 1 \][/tex]
This implies:
[tex]\[ -1 < r < 1 \][/tex]
Next, we use the given expression for [tex]\( a \)[/tex]:
[tex]\[ a = 5(1 - r) \][/tex]
We will evaluate [tex]\( a \)[/tex] for the extreme values of [tex]\( r \)[/tex].
1. When [tex]\( r = -1 \)[/tex]:
[tex]\[ a = 5(1 - (-1)) \][/tex]
[tex]\[ a = 5(1 + 1) \][/tex]
[tex]\[ a = 5 \times 2 \][/tex]
[tex]\[ a = 10 \][/tex]
2. When [tex]\( r = 1 \)[/tex]:
[tex]\[ a = 5(1 - 1) \][/tex]
[tex]\[ a = 5 \times 0 \][/tex]
[tex]\[ a = 0 \][/tex]
Therefore, the resulting range of [tex]\( a \)[/tex] when [tex]\( -1 < r < 1 \)[/tex] is:
[tex]\[ 0 \leq a < 10 \][/tex]
Hence, the correct option is:
(D) [tex]\( 0 \leq a < 10 \)[/tex]
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Next Problem Statement:
Set of solution for: [tex]\( \left| x + \frac{1}{1} \right| > 7 \)[/tex]
First, rewrite the expression inside the absolute value correctly:
[tex]\[ \left| x + 1 \right| > 7 \][/tex]
The property of absolute values gives us two separate inequalities:
[tex]\[ x + 1 > 7 \qquad \text{or} \qquad x + 1 < -7 \][/tex]
Solve each inequality separately.
1. For [tex]\( x + 1 > 7 \)[/tex]:
[tex]\[ x > 7 - 1 \][/tex]
[tex]\[ x > 6 \][/tex]
2. For [tex]\( x + 1 < -7 \)[/tex]:
[tex]\[ x < -7 - 1 \][/tex]
[tex]\[ x < -8 \][/tex]
Thus, the solution set for the given inequality is:
[tex]\[ x > 6 \quad \text{or} \quad x < -8 \][/tex]
In interval notation, this can be written as:
[tex]\[ x \in (-\infty, -8) \cup (6, \infty) \][/tex]
So, the solution set for [tex]\( \left| x + 1 \right| > 7 \)[/tex] is:
[tex]\[ (-\infty, -8) \cup (6, \infty) \][/tex]
