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3 years ago

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If two angles are both vertical and supplementary, can we determine the angles? Is it possible to be both vertical and complemen

tary? If so, can you determine the angles? Explain how you know.

1 answer:

Lorico [155]3 years ago

3 0

Answer:

i. Two angles can be vertical and supplementary.

ii. Yes, we can determine the angles.

Step-by-step explanation:

When two angles are vertical, it implies that they are equal (with respect to vertical opposite property). While two angles are supplementary if and only if they add up to 180 degrees .

It is not mandatory for the two angles to be beside each other so as to be supplementary. So that in this special case, the only measure of angles that can be both vertical and supplementary is 90 degrees .

This answer is right. Please mark Brainliest!

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Aleks04 [339]

The answer is: $f^{-1} = 4x^2-3,\quad\text{for } x \leq 0$

The inverse of a function $f(x)$ is another function, $f^{-1}(x)$ , with the following property:

$f(f^{-1}(x)) = f^{-1}(f(x)) = x$

In other words, the inverse of a function does exactly "the opposite" of what the original function does, and so if you compute them both in sequence you return to the starting point.

Think for example to a function that doubles the input, $f(x)=2x$ , and one that halves it: $f(x)= \frac{x}{2}$ . Their composition is clearly the identity function $f(x)=x$ , since you consider "twice the half of something", or "half the double of something".

In general, to invert a function $y=f(x)$ , you have to solve the expression for $x$ , writing an expression like $x = g(y)$ . If you manage to do so, then $g$ is the inverse of $f$ .

In your case, you have

$f(x) = y = -\frac{1}{2}\sqrt{x+3}$

Multiply both sides by $-2$ to get

$-2y = \sqrt{x+3}$

Square both sides to get

$4y^2 = x+3$

Finally, subtract 3 from both sides to get

$x = 4y^2 - 3$

Since the name of the variables doesn't really have a meaning, you can say that the inverse function is

$f^{-1}(x) = 4x^2 - 3$

As for the domain of the inverse function, remember what we said ad the beginning: if the original function goes from set A (domain) to set B (codomain), then the inverse function goes from set B (domain) to set A (codomain). This means that the inverse function is defined on an element in B if and only if that element belongs to the range of the original function, i.e. the set of the elements of the codomain $b \in B$ such that there exists $a \in A : f(a)=b$ . So, we need the range of $f(x)$ .

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