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kaheart [24]
1 year ago
12

Simplify PLEASE HELP ASAP

Mathematics
2 answers:
lilavasa [31]1 year ago
5 0

\qquad\qquad\huge\underline{{\sf Answer}}

Let's evaluate ~

\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 6x + 9 }

\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 3x  - 3x+ 9 }

\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{}(x - 3) - 3(x - 3) }

\qquad \sf  \dashrightarrow \: \sqrt{ (x - 3)(x - 3) }

\qquad \sf  \dashrightarrow \: \sqrt{ (x - 3) {}^{2}  }

\qquad \sf  \dashrightarrow \: { (x - 3) {}^{}  }

Now, we have been given that value of x :

\qquad \sf  \dashrightarrow \:x < 3

So, let's plug the value of x as 3 in the given expression ~

\qquad \sf  \dashrightarrow \:3 - 3

\qquad \sf  \dashrightarrow \:0

Therefore, we can conclude that :

\qquad \sf  \dashrightarrow \: \sqrt{ {x}^{2} - 6x + 9 }  < 0

Value of the expression should be less than 0

MariettaO [177]1 year ago
3 0

Answer:

0 < x < 6

Step-by-step explanation:

This question asks to solve an inequality which can be written as:

\sqrt{x^2-6x+9} < 3

Solve for x:

\sqrt{x^2-6x+9} < 3\\x^2-6x+9 < 9\\x^2-6x < 0\\x(x-6) < 0

From here we can deduce that the inequality is 0 when:

x = 0 or x = 6

We can write the solution as:

0 < x < 6

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