Answer:
d) one solution; (4, 1)
Step-by-step explanation:
It often works well to follow problem directions. A graph is attached, showing the one solution to be (4, 1).
_____
You know there will be one solution because the lines have different slopes. Each is in the form ...
y = mx + b
where m is the slope and b is the y-intercept.
The first line has slope -1 and y-intercept +5; the second line has slope 1 and y-intercept -3. The slope is the number of units of "rise" for each unit of "run", so it can be convenient to graph these by starting at the y-intercept and plotting points with those rise and run from the point you know.
Answer:
12
Step-by-step explanation:
Length=4
Width=6
Triangle area formula: Length*Width/2
So, 6*4/2
Equal to 6*2
Answer: 12
I think it is Adjacent and Supplementary
Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,
which is a cubic polynomial in 
with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).
Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.
f'(x) = 3x² - 1 = 0 ⇒ x = ±1/√3
So, we have three subsets over which f(x) can be considered invertible.
• (-∞, -1/√3)
• (-1/√3, 1/√3)
• (1/√3, ∞)
By the inverse function theorem,
where f(a) = b.
Solve f(x) = 2 for x :
x³ - x + 2 = 2
x³ - x = 0
x (x² - 1) = 0
x (x - 1) (x + 1) = 0
x = 0 or x = 1 or x = -1
Then 
can be one of
• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);
• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or
• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)
The last one Does not have an awnser there is no like terms
