It's a trick question. There are an infinite number of mixed numbers between 3 and 4 that can multiply to equal 12 (for example, 3 and 3/7 times 3 and 1/2), but there are no mixed numbers between 3 and 4 that can multiply to equal 9. 3 times 3 is not between them but is 3, but that quantity is excluded because 3<x<4. Anything even a small bit above the number 3 would have to be multiplied by 2 and some fraction, which would not be between 3 and 4.
Answer:
M = -2
Step-by-step explanation:
-ab-a=-(-4)(-5)-(-4)=-20+4=-16
so the answer is b
The x-int is (16,0)
you plug in 0 for y and solve
x-6(0)=16
x=16
the y-int is (0, -8/3)
you plug in 0 for x and solve
0-6y=16
y= -16/6
simplify by dividing both the numerator and the denominator by 2
y= -8/3
R²-6r-9s²+9 =
r(r-6) - 9(s²-1)
