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arlik [135]
2 years ago
8

Is it necessary to perform the horizontal line test when finding the inverse of every function?

Mathematics
1 answer:
rewona [7]2 years ago
8 0

Yes, the horizontal line test is important when finding the inverse of a function because without the intersection of the line, they would be no inverse.

<h3>Steps in finding inverse of a function</h3>

The steps include;

  • Replace f(x)with y in the equation of the function
  • Interchange x and y in the function
  • Then solve for y
  • Replace y with f-1(x) in the function

A property of horizontal line is that if it intersects the graph of a function more that on way, then 'f' does not have an inverse.

If no horizontal line intersects the graph , 'then f' does have an inverse.

Thus, the horizontal line test is important when finding the inverse of a function because without the intersection of the line, they would be no inverse.

Learn more about horizontal line test here:

brainly.com/question/11471811

#SPJ1

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n^2-2n-3 where n is the number of key rings in thousands, find the number of key rings sold when the profits is $5,000
ArbitrLikvidat [17]

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For the profit of $5,000, the key chains made is n = 4,000.

Step-by-step explanation:

Here the given expression , where: n = Number of key rings in thousands is given as:

P(n)  =   n²- 2 n - 3

Now, Profit is given to be $5,000.

Also, as we know 5000 = 5  x (1,000)

⇒  n²- 2 n - 3 = 5

Now, solving the above expression for the value of n, we get:

n^2 -  2 n - 3 = 5\\\implies  n^2  -2n - 8  =  0\\\implies  n^2  -4n +  2n - 8  =  0\\\implies n(n-4)  + 2(n-4)  = 0\\\implies (n-4) (n+2)  = 0

So, n = 4, OR , n = -2

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3 years ago
Find the Volume of the cone​
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Answer:

<h2>1005.12</h2>

Step-by-step explanation:

<u>first</u><u> </u><u>we </u><u>need</u><u> to</u><u> </u><u>know</u><u> that</u><u> </u><u>r</u><u>edious</u><u> </u><u>isn't</u><u> </u><u>gien</u>

so we have to find redious

according to Pythagoras theorem

a²+b²=c²

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as \: we \: khow \: volume \: of \: a \: cone

=  \frac{1}{3} \pi {r}^{2} h

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3 years ago
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