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
10,500 mosquitos because 1,000•10hr=10,000 then half of a thousand is 500 so it's that
I hope this helps you
x+11=5x-5
4x=16
x=4
Answer:
It is a right triangle
Step-by-step explanation:
Information needed:
Formula: a^2+b^2= c^2
a: leg
b: leg
c: hypotenuse
the longest side is always the hypotenuse, so 17 in
the order of legs don't matter so 8 in and 15 in
Solve:
a^2+b^2= c^2
8^2+15^2= 17^2
64+225= 289
289= 289
Final answer:
It is a right triangle
Yes, because a rational number is any number that can be expressed as a fraction a/b where a/b are both integers, but cannot be zero.
