Simplify the expression (2 − 5)(3 + 7) − (4 + 3). Show your work!
2 answers:
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Answer:
See below
Step-by-step explanation:
Remember the quotient identity , so you need to take the reciprocal of the sine value for the given angle in order to get the cosecant value for that same angle.
Before we find 
, let's think for a minute about what 
does, and what it means to find the inverse of a function. A function essentially takes a value <em />
from the <em>domain</em> and maps it to another value <em />
in that function's <em>range</em> according to a set of rules. Here, those rules are defined by the formula
All the inverse does is <em>swap the domain and range of the function</em>. Now, instead of trying to map to , we're trying to find a set of rules that'll map back to . To find those rules, all we have to do is solve the above equation for .
First, we'll multiply both sides of the equation by
to get it out of the denominator:
Cancelling on the right side and distributing on the left, we get:
Next, we collect all of our x terms on one side, and all of our non-x terms on the other:
Let's rearrange the right side and factor out an x on the left:
And finally, we divide both sides by
to obtain our answer:
![[x(3y-4)]/(3y-4)=(y+2)/(3y-4)\\\\x= \frac{y+2}{3y-4}](https://tex.z-dn.net/?f=%5Bx%283y-4%29%5D%2F%283y-4%29%3D%28y%2B2%29%2F%283y-4%29%5C%5C%5C%5Cx%3D%20%5Cfrac%7By%2B2%7D%7B3y-4%7D%20)
This equation gives us the "rules" for mapping any given y in the range back to an x in the domain. If we swap the domain and the range, we can define our function
All the inverse does is <em>swap the domain and range of the function</em>. Now, instead of trying to map
First, we'll multiply both sides of the equation by
Cancelling on the right side and distributing on the left, we get:
Next, we collect all of our x terms on one side, and all of our non-x terms on the other:
Let's rearrange the right side and factor out an x on the left:
And finally, we divide both sides by
This equation gives us the "rules" for mapping any given y in the range back to an x in the domain. If we swap the domain and the range, we can define our function