We will assume A×B = B×A and show that A and B are necessarily the same.
Assume A×B = B×A. Let a ∈ A and b ∈ B. Then (a,b) ∈ A×B. Since A×B = B×A, we have (a,b) ∈ B×A. That is, a ∈ B and b ∈ A. Therefore, a ∈ A implies a ∈ B and b ∈ B implies b ∈ A. So A = B, as we had set out to do.
Well first, you have to simplify the equation: which is: y=6x-2
After that, you have to determine the domain, and most of the time, when you can determine the domain, you have to imagine the graph, and in which you can only find the range that way. In order to find the domain however, you just have to know what restricts x. Like in this problem, there are no numbers that violate the rules of math, and since this equation is a linear problem, the domain and range are both ALL REAL NUMBERS.
Answer:
X intercept: (-1.2,0) Y intercept: (0,-0.7)
Step-by-step explanation:
Intercepts are where the line will cross the graph. When figuring out the intercept, the other point will always be 0.
Answer: 1/30
Step-by-step explanation:
∫[0,4] arcsin(x/4) dx = 2π-4
x = 4sin(u)
arcsin(x/4) = arcsin(sin(u)) = u
dx = 4cos(u) du
∫[0,4] 4u cos(u) du
∫[0,4] f(x) dx = ∫[0,π/2] g(u) du
v = ∫[1,e] π(R^2-r^2) dx
where R=2 and r=lnx+1
v = ∫[1,e] π(4-(lnx + 1)^2) dx
Using shells dy
v = ∫[0,1] 2πrh dy
where r = y+1 and h=x-1=e^y-1
v = ∫[0,1] 2π(y+1)(e^y-1) dy
v = ∫[0,1] (x-x^2)^2 dx = 1/30
Answer:
Step-by-step explanation:
we can take common factor of -1 in the denominator
and now multiply and divide by -1
