The range of the equation is 
Explanation:
The given equation is 
We need to determine the range of the equation.
<u>Range:</u>
The range of the function is the set of all dependent y - values for which the function is well defined.
Let us simplify the equation.
Thus, we have;

This can be written as 
Now, we shall determine the range.
Let us interchange the variables x and y.
Thus, we have;

Solving for y, we get;

Applying the log rule, if f(x) = g(x) then
, then, we get;

Simplifying, we get;

Dividing both sides by
, we have;

Subtracting 7 from both sides of the equation, we have;

Dividing both sides by 2, we get;

Let us find the positive values for logs.
Thus, we have,;


The function domain is 
By combining the intervals, the range becomes 
Hence, the range of the equation is 
It’s 42 but Carrier the 2
-π/2 < arctan(x) < π/2
So cos(π/2) < cos(arctan(x)) < cos(0)
0 < cos(arctan(x)) < 1
Answer:
x=36
y=9
Step-by-step explanation:
Plug in 0 for x to find the y-intercept
0 + 4y = 36
4y = 36
y = 9
y-intercept (0, 9)
Plug in 0 for y to find the x-intercept
x + 4(0) = 36
x = 36
x-intercept (36, 0)
10x-6y
You have to add the like terms.