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muminat
3 years ago
14

Simplify. .

Mathematics
1 answer:
garri49 [273]3 years ago
6 0

Answer:

6y+4

Step-by-step explanation:

-(2x-3y) +5x +2 -3x +3y +2

(-2x -3x +5x) (3y +3y)  (2 +2)

0 +6y +4

6y +4

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Which of these numbers are irrational?<br>see in picture for answer choices ​
Likurg_2 [28]

Option A:

\sqrt{17}, \sqrt[3]{14} are irrational numbers.

Solution:

Given numbers:

$\sqrt{17} , 0, \sqrt[3]{8} , \frac{22}{7} , \sqrt{81} , \sqrt[3]{14}

To which of these are irrational numbers:

<u>Irrational:</u>

An irrational number cannot be written in the form \frac{a}{b}, where a and b are integers and b is non-zero.

(1) \sqrt{17} cannot be written as \frac{a}{b}.

So, it is a irrational.

(2) \sqrt[3]{8} =\sqrt[3]{2^3}

Cube and cube roots are get canceled.

\sqrt[3]{8} =2

2 is an integer. So it is not a irrational number.

(3) \frac{22}{7} is in the form \frac{a}{b}.

So it is not a irrational number.

(4)\sqrt{81}=\sqrt{9^2}

Square and square root are get canceled.

\sqrt{81}=9

9 is an integer. So it is not a irrational number.

(5) \sqrt[3]{14}

14 cannot be written as \frac{a}{b} form.

So, it is a irrational number.

Therefore \sqrt{17}, \sqrt[3]{14} are irrational numbers.

Option A is the correct answer.

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7 0
3 years ago
Read 2 more answers
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