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:
1000 times
First, we estimated the Sun's width in terms of the Earth's breadth, then the Star's width in terms of the Earth's breadth, and for comparison, we did simple division, which revealed that the Star KW Sagittarii is 1000 times wider than the Sun.
The distance: use absolute value
Between x and 8: x-8
The distance between x and 8: | x-8 |
“is no more than” means it can be that or be less than that, so use ≤
The distance between x and 8 is no more than 4
| x-8 | ≤ 4
Answer:
where 
denote arc lengths of two circles
Step-by-step explanation:
Let denote arc lengths of two circles, 
denote corresponding radii and
denote the corresponding central angles.
So,
and 
This implies 
and 
As each circle has an arc where the measures of the corresponding central angles are the same, 
As radius of one circle is twice the radius of the other circle,
