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klio [65]
3 years ago
6

What is the solution to the equation? x = 1 x = 6 x = 12 x = 24

Mathematics
1 answer:
stiks02 [169]3 years ago
4 0

Answer:

<h2>X=6</h2>

Option B is the correct option

solution,

\sqrt{5x - 7}  =  \sqrt{3x + 5}

Cancel the square roots on both sides

5x - 7 = 3x + 5

Add 7 to both sides

5x - 7 + 7 = 3x + 5 + 7

Simplify

5x = 3x + 12

Subtract 3x from both sides

5x - 3x = 3x + 12 - 3x

Simplify

2x = 12

Divide both sides by 2

\frac{2x}{2}  =  \frac{12}{2}

Simplify

x = 6

Hope this helps...

Good luck on your assignment..

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Answer: Choice B  y \le 0

========================================================

Explanation:

They want to know what are all the possible y values here. The range is the set of all possible outputs of a function.

The largest y value is y = 0 which occurs at the vertex, in this case the highest point. There is no smallest y value as those arrows point downward forever.

So y is between negative infinity and 0, which means we write -\infty < y \le 0

Converting that to interval notation leads to (-\infty, 0]. Note the square bracket to include the endpoint.

The compound inequality -\infty < y \le 0 simplifies to y \le 0

One way to describe this particular relation's range is to say "The set of negative real numbers with 0 included".

Side note: The domain is the set of all possible x inputs. In this case, the domain is the set of all real numbers.

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\left[\begin{array}{ccc|c}1&2&1&-3\\2&1&-3&-20\\1&-1&5&19\end{array}\right]

Add -2(row 1) to row 2, and -1(row 1) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&-3&4&22\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&9&36\end{array}\right]

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\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&-5&-14\\0&0&1&4\end{array}\right]

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\left[\begin{array}{ccc|c}1&2&1&-3\\0&-3&0&6\\0&0&1&4\end{array}\right]

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So we have \boxed{x=-3,y=-2,z=4}.

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