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Answer :
Sure! Let's solve the inequality step-by-step:
We are given the inequality:
[tex]\[
\frac{1}{3} n + 4.6 \leq 39.1
\][/tex]
and we need to find the possible values for the number [tex]\( n \)[/tex].
1. Subtract 4.6 from both sides:
To isolate the term with [tex]\( n \)[/tex], start by subtracting 4.6 from both sides:
[tex]\[
\frac{1}{3} n \leq 39.1 - 4.6
\][/tex]
2. Simplify the right side:
Perform the subtraction on the right side:
[tex]\[
\frac{1}{3} n \leq 34.5
\][/tex]
3. Solve for [tex]\( n \)[/tex]:
To solve for [tex]\( n \)[/tex], multiply both sides by 3 to get rid of the fraction:
[tex]\[
n \leq 34.5 \times 3
\][/tex]
4. Calculate the result:
Now calculate the multiplication:
[tex]\[
n \leq 103.5
\][/tex]
So, the possible values for the number [tex]\( n \)[/tex] are those that are less than or equal to 103.5. Therefore, the correct answer is [tex]\( n \leq 103.5 \)[/tex].
We are given the inequality:
[tex]\[
\frac{1}{3} n + 4.6 \leq 39.1
\][/tex]
and we need to find the possible values for the number [tex]\( n \)[/tex].
1. Subtract 4.6 from both sides:
To isolate the term with [tex]\( n \)[/tex], start by subtracting 4.6 from both sides:
[tex]\[
\frac{1}{3} n \leq 39.1 - 4.6
\][/tex]
2. Simplify the right side:
Perform the subtraction on the right side:
[tex]\[
\frac{1}{3} n \leq 34.5
\][/tex]
3. Solve for [tex]\( n \)[/tex]:
To solve for [tex]\( n \)[/tex], multiply both sides by 3 to get rid of the fraction:
[tex]\[
n \leq 34.5 \times 3
\][/tex]
4. Calculate the result:
Now calculate the multiplication:
[tex]\[
n \leq 103.5
\][/tex]
So, the possible values for the number [tex]\( n \)[/tex] are those that are less than or equal to 103.5. Therefore, the correct answer is [tex]\( n \leq 103.5 \)[/tex].
Thank you for reading the article The sum of 4 6 and one third of a number is at most 39 1 What are all the possible values of the number. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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Rewritten by : Jeany
