9514 1404 393
Answer:
1) f⁻¹(x) = 6 ± 2√(x -1)
3) y = (x +4)² -2
5) y = (x -4)³ -4
Step-by-step explanation:
In general, swap x and y, then solve for y. Quadratics, as in the first problem, do not have an inverse function: the inverse relation is double-valued, unless the domain is restricted. Here, we're just going to consider these to be "solve for ..." problems, without too much concern for domain or range.
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1) x = f(y)
x = (1/4)(y -6)² +1
4(x -1) = (y-6)² . . . . . . subtract 1, multiply by 4
±2√(x -1) = y -6 . . . . square root
y = 6 ± 2√(x -1) . . . . inverse relation
f⁻¹(x) = 6 ± 2√(x -1) . . . . in functional form
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3) x = √(y +2) -4
x +4 = √(y +2) . . . . add 4
(x +4)² = y +2 . . . . square both sides
y = (x +4)² -2 . . . . . subtract 2
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5) x = ∛(y +4) +4
x -4 = ∛(y +4) . . . . . subtract 4
(x -4)³ = y +4 . . . . . cube both sides
y = (x -4)³ -4 . . . . . . subtract 4
Answer:
x= 6, -4
Step-by-step explanation:
x^2 - 2x - 24 = 0
(x-6) (x+4) = 0
x = 6, x= -4
First, determine if the boundary line should be dotted or solid. In this case, it should be dotted because the symbol is less than not less than or equal to. This leaves you with options B or D. Now, to see if it should be shaded up or down, test it by substituting any point, let's say the origin (0,0), to see if that point is a solution to the equation. If it is, you shade that side of the graph, but if it's not, you shade the other side of the graph.
0 is less than (-3/4)(0) + 2
0 is less than 2.
Because 0 is actually less than 2, the statement is correct and you shade below the line
Answer: D
A:
Area: 289 m
Perimeter: 51 m
B:
Area: 4 m
Perimeter: 8 m
