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Ahat [919]
2 years ago
11

The graph of h(x) is a translation of f (x) = RootIndex 3 StartRoot x EndRoot.

Mathematics
1 answer:
Sergio039 [100]2 years ago
3 0

Transformation involves changing the form of a function. The function that represents is C; g(x) =- \sqrt[3]{x+1}

<h3>What is a function?</h3>

The function is a type of relation, or rule, that maps one input to specific single output.

Function f(x) is given byy;

g(x) = \sqrt[3]{x}

f(x) is reflected across the x-axis.

The rule of this transformation (i.e. reflection) is:

(x, y) ⇒ (x, - y)

So,

f'(x) = -f(x)

This gives

f'(x) = - \sqrt[3]{x}

Now, f'(x) is translated 1 unit left

The rule of this transformation (i.e. translation) is:

(x, y) ⇒ (x+ 1, y)

So,

g(x) = f'(x+ 1)

This gives

g(x) = -\sqrt[3]{x+1}

Hence, the function that represents g(x) is (c)

Learn more about function here:

brainly.com/question/2253924

#SPJ1

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Answer:

Tina is first

Tony is first

Theo is first

Step-by-step explanation:

Events with greater probability are more likely.

Let's put the probabilities in decimal form.

P(Tina is first) = 0.05

P(Theo is first) = %85 = 0.85

P(Tony is first) = 1/10 = 0.1

0.05 < 0.1 < 0.85

The events are listed below from least to most likely.

Tina is first, Tony is first, Theo is first

hope it helps

8 0
3 years ago
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Derivatives of inverse trigonometric functions please explain answers
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I'll explain how to do the first one:-

y = cos-1(x2)

This can be described as ' a function of a function'   x^2 is a function of x and cos-1(x^2) is a function of x^2.

We need to  apply the chain rule.

Personally I find this  easier to understand if i let u = x^2, so

If y = f(u) and u is a function of x then

dy/dx = dy/ du * du/dx

Here u = x^2  and y = cos-1(u)

du/dx = 2x

so dy/dx = d(cos-1(x^2) dx = dy/du * du/dx


= -1 / √(1 - u^2) * 2x

= -2x / √(1 - u^2)    

= -2x / √(1 - (x^2)^2)

= -2x / √(1 - x^4)

I hope this helps. but if not. you might like to employ the formulae in the question - The square boxes contain the 'u' s in my answer. These formulae are equivalent to my explanation.

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