Oh my gosh,
I need an answer for that as well..
Answer: 1. rounds down to 1
2. rounds up to 10
3. rounds down to 7
4. not sure I say it rounds down to 4
5. rounds down to 2
6. rounds down to 6
7. rounds down to 5
8. rounds up to 4
hi!
1) rewrite as a fraction
(2^4/2^2)^3
2) reduce the expression by cancelling out the common factors
so factor out 2^2 out of 2^4
3) multiply by 1 so,
(2^2 * 2^2/ 2^2 * 1)^3
4) continue to cancel out common factors the rewrite the expression to be
(2^2/1)^3
5) then divide 2^2 by 1 so, it will be (2^2)^3
6) lastly simplify the expression so apply the power rule and multiply so it will be 2^2*3
then it will be 2^6
and raise 2 to the power of 6 and you answer is 64
Answer:
a) and d) are bijections. b) and c) are not
Step-by-step explanation:
a) Every linear non constant function is a bijection. We can easily find the inverse of f by making a simple calculus.
If y is on the function image, we have y = -3x + 4 for certain x, then
y- 4 = -3x
-(y-4)/3 = x
therefore
b) -3 * X² + 4 is not a bijection because quadratic funtions arent bijective. If you evaluate in opposite values you will obtain the same result. For example f(-1) = f(1) = 6
c) (x+1)/(x+2) is not a bijection. It isnt even defined in -2 because the denominator is equal to 0 if X= -2 and we cant divide by 0. A bijective function from R to R must be defined in every element of R. In general, homographic non linear functions are not bijective for the same reason this function is not.
d) is bijective. There isnt a simple argument we can use to conclude this. We have no other choice than trying to find the inverse function by making a calculus.
Y =
Y-1 =
Not that since 5 is odd, we can calculate independently of which value Y-1 takes. Therefore , and we can conclude that f is bijective.
I hope this helps you!