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2 years ago

For f(x)=1/x^2-3

Mathematics

1 answer:

igomit [66]2 years ago

5 0

For f(x)=1/x^2-3

Find

A) f(3)

B) f(2-h)

If f(x)=1/x^2-3, then f(3) = 1 / 3^2 - 3. The exponentiation here must be carried out first: f(3) = 1/9 - 3. Then f(3) = 1/9 - 27/9 = -26/9

If f(x)=1/x^2-3, then f(2-h) = 1 / [2-h]^2 - 3. This result may be left as is or expanded. In expanded form, we have:

f(2-h) = ------------------ - 3

4-4h +h^2

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we know that

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m∠ZXY=37°-----> given value

m∠WXY=m∠ZXY=37°

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the answer part a)

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We start with the expression:

$\sqrt{y^3} + \sqrt{16*y^3} - 4*y\sqrt{y}$

where y > 0. (this allow us to have y inside a square root, so we don't mess with complex numbers)

We want to find the equivalent expression to this one.

Here, we can do the next two simplifications:

$\sqrt{16*y^3} = \sqrt{16} \sqrt{y^3} = 4*\sqrt{y^3}$

And:

$y*\sqrt{y} = \sqrt{y^2} *\sqrt{y} = \sqrt{y^2*y} = \sqrt{y^3}$

If we apply these two to our initial expression, we can rewrite it as:

$\sqrt{y^3} + \sqrt{16*y^3} - 4*y\sqrt{y}$

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Here we can use the second simplification again, to rewrite:

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Then the correct option is C.

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