<span>Simplifying
6x * 3y * 9x2y4
Reorder the terms for easier multiplication:
6 * 3 * 9x * y * x2y4
Multiply 6 * 3
18 * 9x * y * x2y4
Multiply 18 * 9
162x * y * x2y4
Multiply x * y
162xy * x2y4
Multiply xy * x2y4
162x3y<span>5</span></span>
Answer:
29-12 is a 17 difference
52-29 is a 23 difference (adding 6 to the number 17)
81-52 is a 29 difference (adding 6 to the number 23)
116-81 is a 35 difference (adding 6 to the number 29)
Step-by-step explanation:
For quadratics with an a value (coefficient of x²) of 1, we find factors of c, the constant, that sum to b, the coefficient of x. In our case we want factors of -16 that sum to 6.
-16 = -1(16) or 16(-1)
-16 = -2(8) or 8(-2)
-16 = -4(4) or 4(-4)
Out of these 6 possibilities, the only one that sums to 6 is 8(-2), because 8+(-2)=6.
This gives us the factors (x+8)(x-2).
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!
