Answer:
The maximum speed of Hal's airplane in still air is:
The speed of the wind
Step-by-step explanation:
Remember that the velocity v equals the distance d between time t.
and 
The distance that Hal travels when traveling with the wind is:
miles
Where v is the speed of Hal and c is the wind speed.
The distance when traveling against the wind is:
miles
Now we solve the first equation for v
Now we substitute the value of v in the second equation and solve for c
Then:
The maximum speed of Hal's airplane in still air is:
The speed of the wind
F(x)......ur output = 19
19 = 2x^2 + 3x + 5
2x^2 + 3x + 5 - 19 = 0
2x^2 + 3x - 14 = 0
(2x + 7)(x - 2) = 0
2x + 7 = 0
2x = -7
x = -7/2
x - 2 = 0
x = 2
I believe there is 2 possible answers......x (ur input value) = -7/2 or 2
25/100=1/4
a) you bought 1 1/4 pound of ham
b) $6.50
1.25 x $5.20=$6.50
Answer:
34a
Step-by-step explanation:
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.
__
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
