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nirvana33 [79]
3 years ago
7

John rode 2 kilometers on his bike and his sister rode 3000 meters who rode the farthest and how much farther did they ride.(ans

wer in KM)
Mathematics
2 answers:
user100 [1]3 years ago
7 0
John 2km
sister 3km
his sister rode the fastest
svp [43]3 years ago
7 0
His sister rode the farthest by 1km 
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Will someone explain what i did wrong here?
tensa zangetsu [6.8K]

The actual inverse function is:

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

And the domain is [0, ∞).

<h3>Where is the mistake?</h3>

Remember that for a given function f(x) with a domain D and a range R.

For the inverse function, f⁻¹(x) the domain is R and the range is D.

Here, for the given function the domain is x ≥ 3 and the range is [0, ∞).

Then for the inverse function, which is:

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

(to check this, you must have that):

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

The domain will be [0, ∞) and the range x ≥ 3

If you want to learn more about inverse functions:

brainly.com/question/14391067

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1 year ago
I need help please???
alexdok [17]

Answer:

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Step-by-step explanation:

This would be the BEST answer. Hope it helps! ☺

4 0
2 years ago
The graph shows two linear functions, f and g. Which formula BEST represents g(x)?
Softa [21]

Answer: C

g(x) = 1/2 f (x)

Step-by-step explanation:

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3 years ago
What is the value of x, if the perimeter of the rectangle is 42? *<br><br> This is the problem :
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Answer:

Im assuming you meant 43 because if you did x=5

Step-by-step explanation:

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5 0
2 years ago
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Simplify the trigonometric expression.
Natasha_Volkova [10]
First we are going to find the common denominator of both fractions. To do that, we are going to multiply their denominators:
(1+sin \alpha )(1-sin \alpha )=1-sin^2 \alpha

Now we can rewrite our expression using the common denominator:
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Finally, we can use the trig identities: 1-sin^2 \alpha =cos^2 \alpha and sec \alpha = \frac{1}{cos \alpha } to simplify our trig expression:
\frac{2}{cos^2\alpha}=2sec^2 \alpha

We can conclude that the correct answer is the fourth one.
5 0
3 years ago
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