Just sub in (x+1) wherever x appears:
f(x) = 2x+4
f(x+1) = 2(x+1)+4 = 2x + 2 + 4 = 2x+6
Here is your answer
The square root of a squared number, will be the number itself.
EXPLANATION:
Let
be a number.
Its square= ![{x}^{2}](https://tex.z-dn.net/?f=%7Bx%7D%5E%7B2%7D)
Now square root of this squared number= ![\sqrt{{x}^{2}}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%7Bx%7D%5E%7B2%7D%7D%20)
= ![{x}^{2×\frac{1}{2}}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%C3%97%5Cfrac%7B1%7D%7B2%7D%7D%20)
= ![{x}^{1}](https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B1%7D%20)
= ![x](https://tex.z-dn.net/?f=%20x%20)
Hence, the initial number x is obtained.
HOPE IT IS USEFUL
Answer:
y=2 is correct
Step-by-step explanation:
as you go across the y axis the points go up by 2
The answer for this question is 25
Answer:
![\large\boxed{\dfrac{4}{x}+6}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Cdfrac%7B4%7D%7Bx%7D%2B6%7D)
Step-by-step explanation:
Six more than the quotient of four and a number x:
![\dfrac{4}{x}+6](https://tex.z-dn.net/?f=%5Cdfrac%7B4%7D%7Bx%7D%2B6)