Find the GCF of 18 and 11.
2 answers:
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Answer:
The functions are inverses; f(g(x)) = x ⇒ answer D
⇒ answer D
Step-by-step explanation:
* <em>Lets explain how to find the inverse of a function</em>
- Let f(x) = y
- Exchange x and y
- Solve to find the new y
- The new y =
* <em>Lets use these steps to solve the problems</em>
∵
∵ f(x) = y
∴
- Exchange x and y
∴
- Square the two sides
∴ x² = y - 3
- Add 3 to both sides
∴ x² + 3 = y
- Change y by
∴
∵ g(x) = x² + 3
∴
∴ <u><em>The functions are inverses to each other</em></u>
* <em>Now lets find f(g(x))</em>
- To find f(g(x)) substitute x in f(x) by g(x)
∵
∵ g(x) = x² + 3
∴
∴ <u><em>f(g(x)) = x</em></u>
∴ The functions are inverses; f(g(x)) = x
* <em>Lets find the inverse of h(x)</em>
∵ h(x) = 3x² - 1 where x ≥ 0
- Let h(x) = y
∴ y = 3x² - 1
- Exchange x and y
∴ x = 3y² - 1
- Add 1 to both sides
∴ x + 1 = 3y²
- Divide both sides by 3
∴
- Take √ for both sides
∴ ±
∵ x ≥ 0
∴ We will chose the positive value of the square root
∴
- replace y by
∴
Answer:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221
Step-by-step explanation:
For this case first we need to create the sample of size 20 for the following distribution:
And we can use the following code: rnorm(20,50,6) and we got this output:
> a<-rnorm(20,50,6)
> a
[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905
[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501
Then we can find the mean and the standard deviation with the following formulas:
> mean(a)
[1] 50.72451
> sqrt(var(a))
[1] 7.470221