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3 years ago

What is the inverse of f(x)=ln(x-6)

1 answer:

6 0

To find the inverse of a function, simply switch the x and f(x). Then make it so that the equation is set equal to f(x). Also note that ln is the log of e:

f(x) = ln(x - 6)

x = ln(f(x) - 6)

e^x = f(x) - 6

$f^{-1} ( x) = e^{x} +6$

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Answer:

The product is 1.4993772e+15.

Step-by-step explanation:

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3 years ago

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strojnjashka [21]

Answer:

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Step-by-step explanation:

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3 years ago

Find the sum of the following infinite geometric series, if it exists. 2 + 6 + 18 + 54 +… Does not exist 23,567 25,982 29,034

Sladkaya [172]

Option A: The sum for the infinite geometric series does not exist

Explanation:

The given series is $2+6+18+54+.......$

We need to determine the sum for the infinite geometric series.

<u>Common ratio:</u>

The common difference for the given infinite series is given by

$r=\frac{6}{2}=3$

Thus, the common difference is $r=3$

<u>Sum of the infinite series:</u>

The sum of the infinite series can be determined using the formula,

$S_{\infty}=\frac{a}{1-r}$ where $0$

Since, the value of r is 3 and the value of r does not lie in the limit $0$

Hence, the sum for the given infinite geometric series does not exist.

Therefore, Option A is the correct answer.

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3 years ago

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3 years ago

An arithmetic progression is a sequence of numbers in which the distance (or difference) between any two successive numbers if t

earnstyle [38]

Answer:

Step-by-step explanation:

Solution:

- We are to write a program for evaluating the sum to Nth of an arithmetic sequence such that the sequence starts from positive integer 1, 3 , 5 , 7 , .. n.

- The sum to nth for the arithmetic series is given by two parameters i.e first integer a = 1 and the distance between successive integers d = 2 in our case.

- For any general distance d we can write our sum to nth as:

Sum to nth = a + (a+d) + (a+2*d) + (a+3*d) .... (a + (n-1)*d)

- From above sequence we can see that every successive number is increased by distance d and added in previous answer.

- We will use an iteration loop for a variable "sum", which is cycled by a "range ( , , )" function.

- The parameters of the range functions corresponds to:

range ( first integer , last integer , step size )

range ( a , n + 1 , d )

- Then we can cast the loop as follows:

" int sum = 0

int d = 2

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for i in range ( a , n + 1 , d )

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- We see that iteration parameter i starts from a = 1, with step size d = 2 and the sum is previously stored sum value plus i for the current loop.

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3 years ago

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