52
100 percent sure though
Answer:
Step-by-step explanation:
the transverse axis is horizontal.
so its a horizontal hyperbola
Center is the origin so center is (0,0)
Equation of horizontal hyperbola is
Given a= 55000 and c= 81000
c^2 = a^2 + b^2
81000^2 = 55000^2 + b^2
subtract 55000^2 on both sides
b = sqrt(81000^2 - 55000^2)= 59464.27
now plug in the values
Linear combination method just means to add the equations so that one variable with cancel. In this case the a values are already set up so that they will cancel.
2a + 3b = 1
-2a + b = 11
4b = 12
b = 3
now put 3 back into either of the original equations and solve for a.
2a + 3(3) = 1
2a + 9 = 1
2a = -8
a = -4
(-4,3)
Problem 1
<h3>Answer: False</h3>
---------------------------------
Explanation:
The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.
So,
f(x) = x+1
f( g(x) ) = g(x) + 1 .... replace every x with g(x)
f( g(x) ) = 6x+1 ... plug in g(x) = 6x
(f o g)(x) = 6x+1
Now let's flip things around
g(x) = 6x
g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)
g( f(x) ) = 6(x+1) .... plug in f(x) = x+1
g( f(x) ) = 6x+6
(g o f)(x) = 6x+6
This shows that (f o g)(x) = (g o f)(x) is a false equation for the given f(x) and g(x) functions.
===============================================
Problem 2
<h3>Answer: True</h3>
---------------------------------
Explanation:
Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.
For example, let
f(x) = 1/(x+2)
g(x) = -2
The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.
So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).
