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valkas [14]
3 years ago
15

PLEASE HELP The Pythagorean Theorem Determine the algebraic representation of f(x)

Mathematics
1 answer:
tiny-mole [99]3 years ago
7 0
<h3>Answer for the first box = 1</h3><h3>Answer for the second box = 2</h3><h3>Answer for the third box = 0</h3>

This gives f(x) = 1\sqrt{x-2}+0 which is the same as  f(x) = \sqrt{x-2}

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

Explanation:

The general template for a square root function is

y = a\sqrt{x-h}+k

where,

  • 'a' handles the vertical stretching/compressing of the graph
  • h determines the horizontal shifting
  • k determines the vertical shifting

The graph has not been scaled vertically. In other words, it has not been vertically stretched or vertically compressed. So the value of 'a' is equal to 1.

The value of h is 2 since the graph has been shifted 2 units to the right. If we wanted to go left, then h would be negative.

The value of k is 0 because there is no vertical shifting going on.

In summary,

a = 1, h = 2, k = 0

Leading

y = a\sqrt{x-h}+k

to turn into

y = 1\sqrt{x-2}+0

which simplifies to

y = \sqrt{x-2}

The Pythagorean theorem is not used in this problem.

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Answer:

\left[\begin{array}{cc}-\frac{1}{2}&\frac{1}{6}\\-\frac{1}{2}&\frac{1}{3}\\\end{array}\right]

Step-by-step explanation:

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<em />

<em>Using the formula, let's find the inverse:</em>

<em>\frac{1}{ad-bc}\left[\begin{array}{cc}d&-b\\-c&a\\\end{array}\right]\\=\frac{1}{(-4)(6)-(-6)(2)}\left[\begin{array}{cc}6&-2\\6&-4\\\end{array}\right]\\=\frac{1}{-12}\left[\begin{array}{cc}6&-2\\6&-4\\\end{array}\right]\\=\left[\begin{array}{cc}-\frac{1}{2}&\frac{1}{6}\\-\frac{1}{2}&\frac{1}{3}\\\end{array}\right]</em>

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