I believe that the answer is All integers where n
![\geq](https://tex.z-dn.net/?f=%20%5Cgeq%20)
1
Answer: Choice A
g(x) = sqrt(2x)
====================================================
Explanation:
"sqrt(x)" is shorthand for "square root of x"
f(x) = 3x^2 is given
g( f(x) ) = x*sqrt(6) is also given
One way to find the answer is through trial and error. This would only apply of course if we're given a list of multiple choice answers.
------------------
Let's start with choice A
g(x) = sqrt(2x)
g( f(x) ) = sqrt(2 * f(x) ) .... replace every x with f(x)
g( f(x) ) = sqrt(2 * 3x^2 ) .... plug in f(x) = 3x^2
g( f(x) ) = sqrt(6x^2 )
g( f(x) ) = sqrt(x^2 * 6)
g( f(x) ) = sqrt(x^2)*sqrt(6)
g( f(x) ) = x*sqrt(6)
We found the answer on the first try. So we don't need to check the others.
------------------
But let's try choice B to see one where it doesn't work out
g(x) = sqrt(x + 3)
g( f(x) ) = sqrt( f(x) + 3)
g( f(x) ) = sqrt(3x^2 + 3)
and we can't go any further other than maybe to factor 3x^2+3 into 3(x^2+1), but that doesn't help things much to be able to break up the root into anything useful. We can graph y = x*sqrt(6) and y = sqrt(3x^2 + 3) to see they are two different curves, so there's no way they are equivalent expressions.
(x-4)² + (x-2)² = x²
(x² - 8x + 16 ) + (x² - 4x + 4) = x
2x² -12x + 20 = x
2x²-12x + 20 - x = 0
2x² - 13x + 20 = 0
(2x-5) (x-4) = 0
2x-5 = 0
2x = 5
x = ⁵/₂
x-4 = 0
x = 4
so x₁ = ⁵/₂ , and x₂ = 4
Answer:
The function which represents this sequence will be:
![a_n=-15n+110](https://tex.z-dn.net/?f=a_n%3D-15n%2B110)
Hence, option (A) is true.
Step-by-step explanation:
Given the sequence
![95, 80, 65, 50, ...](https://tex.z-dn.net/?f=95%2C%2080%2C%2065%2C%2050%2C%20...)
An arithmetic sequence has a constant difference 'd' and is defined by
![a_n=a_1+\left(n-1\right)d](https://tex.z-dn.net/?f=a_n%3Da_1%2B%5Cleft%28n-1%5Cright%29d)
computing the differences of all the adjacent terms
![80-95=-15,\:\quad \:65-80=-15,\:\quad \:50-65=-15](https://tex.z-dn.net/?f=80-95%3D-15%2C%5C%3A%5Cquad%20%5C%3A65-80%3D-15%2C%5C%3A%5Cquad%20%5C%3A50-65%3D-15)
As the difference is the same, so
![d = -15](https://tex.z-dn.net/?f=d%20%3D%20-15)
as
![a_1=95](https://tex.z-dn.net/?f=a_1%3D95)
Thus, substituting
,
in the nth term of an arithmetic sequence
![a_n=a_1+\left(n-1\right)d](https://tex.z-dn.net/?f=a_n%3Da_1%2B%5Cleft%28n-1%5Cright%29d)
![a_n=-15\left(n-1\right)+95](https://tex.z-dn.net/?f=a_n%3D-15%5Cleft%28n-1%5Cright%29%2B95)
![a_n=-15n+110](https://tex.z-dn.net/?f=a_n%3D-15n%2B110)
Therefore, the function which represents this sequence will be:
![a_n=-15n+110](https://tex.z-dn.net/?f=a_n%3D-15n%2B110)
Hence, option (A) is true.