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3 years ago

Find the value of in the triangle shown below 9, 7, and x around the triangle

Mathematics

1 answer:

enot [183]3 years ago

4 0

Answer:

5.66 units

Step-by-step explanation:

A's per our discussion, x is a leg of right triangle.

Therefore, by Pythagoras theorem:

$x = \sqrt{ {9}^{2} - {7}^{2} } \\ = \sqrt{81 - 49} \\ = \sqrt{32}$

$x = \sqrt{ {16}^{2} - {12}^{2} } \\ = \sqrt{256 - 144} \\ = \sqrt{112}$

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35. Explain the difference between the reciprocal of x^2

Scorpion4ik [409]

Answer:

As you can see, the difference between the reciprocal of $x^2$ and the inverse of $x^2$ is that $\frac{1}{x^2} =x^{-2}$ and $\sqrt{x} =x^{\frac{1}{2}$ .

Step-by-step explanation:

First lets find both the reciprocal of $x^2$ and the inverse of

Recall that the reciprocal of a value is where you take a fraction and swap the places of the terms. In the case of $x^2$ , 1 is the denominator, so

$\frac{x^2}{1} =\frac{1}{x^2}$

To find the inverse of a function, you first need swap the locations of x and y in the equation

$y=x^2\\\\x=y^2$

Now, you need to solve for y

$y^2=x\\\\y=\sqrt{x}$

Now, lets rewrite each of these to better compare them

$\frac{1}{x^2} =x^{-2}\\\\\sqrt{x} =x^{\frac{1}{2}$

As you can see, the difference between the reciprocal of $x^2$ and the inverse of

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