Ten less than twice a number is equal to at most 52. What are all the possible values of the number? Inuk wrote the inequality 2x−10≤52 , where x equals the number, to help solve this problem. Solve his inequality. Use the letter x as your variable and write your x term first.
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The inequality 2x - 10 ≤ 52 can be solved as follows:
2x ≤ 52 + 10
2x ≤ 62
x ≤ 31
Therefore, the possible values of the number are all real numbers less than or equal to 31.
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I apologize for the mistake. Let's solve the inequality correctly.
2x - 10 ≤ 52
Add 10 to both sides:
2x ≤ 52 + 10
2x ≤ 62
Now divide both sides by 2:
x ≤ 62/2
x ≤ 31
So the possible values of the number are all real numbers less than or equal to 31.
where x equals the number, to help solve this problem. Solve his inequality. Use the letter x as your variable and write your x term first.
Apologies for the confusion.
To solve the inequality 2x - 10 ≤ 52, we will isolate the variable x:
2x - 10 ≤ 52
Add 10 to both sides:
2x ≤ 52 + 10
2x ≤ 62
Divide both sides by 2:
x ≤ 62/2
x ≤ 31
Therefore, the value of the number (x) must be less than or equal to 31.
To solve the inequality 2x - 10 ≤ 52 for the variable x, we need to isolate x on one side of the inequality sign.
Step 1: Add 10 to both sides of the equation to eliminate the -10 on the left side:
2x - 10 + 10 ≤ 52 + 10
2x ≤ 62
Step 2: Divide both sides of the equation by 2 to isolate x:
2x/2 ≤ 62/2
x ≤ 31
So, the inequality 2x - 10 ≤ 52 is solved as x ≤ 31.
Now, let's find all the possible values of the number x that satisfy this inequality. Since x ≤ 31, the possible values for x can be any number less than or equal to 31. In other words, the number can take any value from negative infinity up to and including 31.
Therefore, the possible values of the number x are all real numbers such that x ≤ 31.