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:
A $60
Step-by-step explanation:
Answer: A. 
B. A'(5) = 1.76 cm/s
Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.
A. Area of a circle is given by

So to find the rate of the area:


Using 

Then
![\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D2.%5Cpi.r.%5B%5Cfrac%7B726%7D%7B%28t%2B11%29%5E%7B3%7D%7D%5D)
![\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdt%7D%3D2.%5Cpi.%5B3-%5Cfrac%7B363%7D%7B%28t%2B11%29%5E%7B2%7D%7D%5D.%5Cfrac%7B726%7D%7B%28t%2B11%29%5E%7B3%7D%7D)
Multipying and simplifying:

The rate at which the area is increasing is given by expression
.
B. At t = 5, rate is:




At 5 seconds, the area is expanded at a rate of 1.76 cm/s.
8x + 6y = 10 ----------- (1)
5x - 3y = 2.2 ---------- (2)
(2) x 2:
10 x - 6y = 4.4 --------(2a)
(2a) + (1)
18x = 14.4
x = 1.8 ------- sub into (1)
8 (1.8) + 6y = 10
14.4 + 6y = 10
6y = 10 - 14.4
6y = - 4.4
y = - 0.73
There is 1 set of solution to it. (Answer C)
You got the b part wrong. It would be 4B+4C/2=4D