<span>let y = sec^2 ( pi x )
y' = 2 sec ( pi x ) sec( pi x ) tan ( pi x ) pi
y' = 2pi sec^2 ( pi x ) tan ( pi x )
y''= 2pi sec^2 ( pi x ) * sec^2 ( pi x ) * pi + 2pi tan ( pi x ) * 2pi sec^2 ( pi x ) tan ( pi x )
y'' = 2 pi^2 sec^4 ( pi x ) + 4 pi^2 sec^2 ( pi x ) tan^2 ( pi x )</span>
Answer:
it should be the negative repirocal of -2 so 1/2
Answer: The answer is 
Step-by-step explanation: We are given four different functions of the variable 'x' and a graph. We are told to select one of the four options that which function can be graphed as the graph given in the question.
To check, we start plotting the functions one by one on a graph paper. We see that the graph of first three functions do not match with the given graph, but the graph of the fourth function given by
matches exactly with the graph given in the question. The attached figure will show the graph for this function, which is exactly same as given.
Thus, the correct option is
Answer is C!
Ok I got to organize the info...
3/15 and 12/55
3*2...
Aha
3*4 equals 12 but not 15...
A: No
B:
8/24 12/35
No
GAUTIE
Sorry...
It's C
5/18 = 25/90
5*5=25
18*5 = 90
Hope that helps you!
One more happy customer!
<3
Answer:
Step-by-step explanation:
Rearrange each function to solve for x.
Switch x and y,
The resulting equation is the inverse function.
A:
f(x) = y = 5+x
x = y-5
y = x-5
f⁻¹(x) = x-5
g(x) = 5-x ≠ f⁻¹(x)
g(x) is not the inverse of f(x).
:::::
B:
f(x) = y = 2x-9
x = (y+9)/2
y = (x+9)/2
f⁻¹(x) = (x+9)/2
g(x) = (x+9)/2 = f⁻¹(x)
g(x) is the inverse of f(x).
:::::
C:
f(x) = y = 2/x - 6
x = 2/(y+6)
y = 2/(x+6)
f⁻¹(x) = 2/(x+6)
g(x) = (x+6)/2 ≠ f⁻¹(x)
:::::
D:
f(x) = y = x/3 + 4
x = 3y - 12
y = 3x - 12
f⁻¹(x) = 3x - 12
g(x) = 3x - 4 ≠ f⁻¹(x)
g(x) is not the inverse of f(x).
