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Marat540 [252]
3 years ago
15

Find All numbers whose absolute value is 2

Mathematics
1 answer:
Elanso [62]3 years ago
7 0

Answer:

Step-by-step explanation:

Absolute Value. Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative. The absolute value of 2 + 7 is 5.

Has two solutions x = a and x = -a because both numbers are at the distance a from 0. ... An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative.

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schepotkina [342]

Answer:

                            \large\boxed{\large\boxed{\sqrt{3}

Explanation:

You are comparing irrational numbers.

By inspection, i.e. at first sight you can only compare \sqrt{3} \text{ }and\text{ } 2\sqrt{3} because they have the same radicand.

You can order: \sqrt{3}

You can introduce the 2 inside the radical by squaring it:

       2\sqrt{3}=\sqrt{2^2\times3}=\sqrt{12}

Since 5 is between 3 and 12, you can order:

  • \sqrt{3}

Which is:

  • \sqrt{3}

You must know that π ≈ 3.14.

5 is less than 9 and the square root of 9 is 3; hence, \sqrt{5} and \sqrt{5}

Now you must determine whether π is less than or greater than \sqrt{12}

Using a calculator or probing numbers between 3 and 4 you get \sqrt{12} \approx3.46

Hence, the complete order is:

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3 years ago
Do you know the answer to this question
Vika [28.1K]

Answer:

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3 years ago
Michael is packing books into boxes. Each box can hold either 15 small books or
zavuch27 [327]

Let x equal the number of tiny book boxes and y equal the number of large book boxes.

Because John must pack at least 35 boxes:

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x + y ≥ 35

 

Since John needs to pack at least 350 books:

 

(number of small books packed) + (number of large books packed) (is at least) 350

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Your system of inequalities is:

 

x + y ≥ 35

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x^2-6x+6=0\\\\a=1;\ b=-6;\ c=6\\\Delta=b^2-4ac\\\\\Delta=(-6)^2-4\cdot1\cdot6=36-24=12 \ \textgreater \  0\\\\then\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{12}=\sqrt{4\cdot3}=\sqrt4\cdot\sqrt3=2\sqrt3\\\\x_1=\dfrac{6-2\sqrt3}{2\cdot1}=\dfrac{6-2\sqrt3}{2}=\boxed{3-\sqrt3}\\\\x_2=\dfrac{6+2\sqrt3}{2\cdot1}=\dfrac{6+2\sqrt3}{2}=\boxed{3+\sqrt3}
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