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sasho [114]
3 years ago
12

What are the solutions to the absolute value inequality |x − 70| ≤ 3? Remember, the inequality can be written as −3 ≤ x − 70 ≤ 3

or as x − 70 ≤ 3 and x − 70 ≥ −3
Mathematics
2 answers:
irakobra [83]3 years ago
6 0

Answer:

67 ≤ x ≤ 73

Step-by-step explanation:

Given inequality is:  |x − 70| ≤ 3

We have to find the value of x.

If  |x| ≤ a then  x ≤ a and x ≥ -a .

Applying this rule to given inequality,we get  x - 70 ≤ 3  and  x - 70 ≥ -3

Adding 70 to both sides of above both inequality,we get

x-70+70 ≤ 3+70 and x-70+70 ≥ -3+70

Adding like terms,we get

x ≤ 73 and x ≥ 67  

Combining above two inequalities ,we get

67 ≤ x ≤ 73  which is the answer.


ikadub [295]3 years ago
5 0

Answer:

solutions to the absolute value inequality is 67\leq x\leq 73

Step-by-step explanation:

To find the solution of x absolute value of  |x − 70| ≤ 3 will be written in the form of interval because the given fraction (x-70) is less than and equal to 3.

-3\leq(x-70)\leq 3

Now we add 70 on every part of the inequality.

-3+70\leq (x-70)+70\leq 3+70

67\leq x\leq 73

So the solution to the absolute value inequality is 67\leq x\leq 73.


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See explanation

Step-by-step explanation:

Consider the first triangle:

1. From the right triangle with acute angle \theta_1:

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2. From the right triangle with acute angle \alpha:

\dfrac{c+a}{h}=\tan\alpha\Rightarrow c+a=h\tan\alpha

Thus,

h\tan\theta_1+a=h\tan\alpha\Rightarrow a=h\tan\alpha-h\tan\theta_1

Consider the second triangle:

1. From the right triangle with acute angle \theta_2:

\dfrac{d}{h}=\tan\theta_2\Rightarrow d=h\tan\theta_2

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h\tan\theta_2+b=h\tan\beta\Rightarrow b=h\tan\beta-h\tan\theta_2

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h\tan\beta-h\tan\theta_1

Note that this solution is true only for acute angles \alpha,\ \beta,\ \theta_1,\ \theta_2

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