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MAXImum [283]
3 years ago
9

| 3x | = 9 | -3r | = 9 b/5 = 1

Mathematics
1 answer:
Ierofanga [76]3 years ago
8 0
|3x| = 9
|3x| = ±9
|3x| = 9    U    |3x| = -9
  3x = 9     U     3x = -9
   3     3              3      3
    x = 3               x = -3

|-3r| = 9
|-3r| = ±9
|-3r| = 9    U    |-3r| = -9
   3r = 9      U     3r = -9
    3    3               3     3
    r = 3                 r = -3

     ¹/₅b = 1
5(¹/₅b) = 5(1)
       b = 5
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Step-by-step explanation:

So we have the expression:

12-[20-2(6^2\div3\times2^2)]

Recall the order of operations or PEMDAS:

P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.

E: Within the parentheses, if exponents are present, do them before all other operations.

M/D: Multiplication and division next, whichever comes first.

A/S: Addition and subtraction next, whichever comes first.

(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)

Thus, we should do the operations inside the parentheses first. Therefore:

12-[20-2(6^2\div3\times2^2)]

The parentheses is:

(6^2\div3\times2^2)

Square the 6 and the 4:

(36\div3\times4)

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

(36\div3\times4)\\=(12\times4)\\=48

Therefore, the original equation is now:

12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]

Multiply with the brackets:

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Subtract with the brackets:

=12-[-76]

Two negatives make a positive. Add:

=12+76=88

Therefore:

12-[20-2(6^2\div3\times2^2)]=88

3 0
3 years ago
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