(0,0) represents the very beginning of the rental period:  0 time has elapsed, and thus there is no charge.
(1,125) represents this situation 1 hour after the rental has begun, and shows that the amount due at that point for the rental is 
$125.
 
        
                    
             
        
        
        
Answer:
(SS,SG,GS,GG) Probability 0.25
Step-by-step explanation:
i don't know how to do this
 
        
             
        
        
        
Answer:
23x-1
Step-by-step explanation:
 
        
             
        
        
        
Answer:
Step-by-step explanation:
We'll take this step by step.  The equation is
![8-3\sqrt[5]{x^3}=-7](https://tex.z-dn.net/?f=8-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-7)
Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time.  First thing is to subtract 8 from both sides:
![-3\sqrt[5]{x^3}=-15](https://tex.z-dn.net/?f=-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-15)
The goal is to isolate the term with the x in it, so that means that the -3 has to go.  Divide it away on both sides:
![\sqrt[5]{x^3}=5](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E3%7D%3D5)
Let's rewrite that radical into exponential form:

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely.  On the right, let's rewrite that back in radical form to solve it easier:
![x=\sqrt[3]{5^5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E5%7D)
Let's group that radicad into groups of 3's now to make the simplifying easier:
![x=\sqrt[3]{5^3*5^2}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E3%2A5%5E2%7D) because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:
 because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:
![x=5\sqrt[3]{5^2}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B5%5E2%7D) which is the same as:
 which is the same as:
![x=5\sqrt[3]{25}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B25%7D)
 
        
             
        
        
        
<h3>
Answers:</h3>
- A. T <-> U  is a <u>biconditional</u>
- B. (A & B) v (C & D)  is a <u>disjunction</u>
- C. R -> ~S  is a <u>conditional</u>
- D. P & Q   is a <u>conjunction</u>
- E. ~(R v P)  is a <u>negation</u>
========================================
Explanations:
- A biconditional is anything in the form A <-> B. This is a compact way of saying (A -> B) & (B -> A). We replace A and B with logical statements.
- Disjunctions are of the basic form A v B. The "v" basically means "or".
- Any conditional is of the form "if... then...". For example, "if it rains, then it gets wet outside" is a conditional. In terms of logic symbols, we write A -> B to mean "if A, then B". 
- Conjunctions are whenever we combine two logical statements with an "and" or an ampersand symbol. The basic form is A & B
- Negations are the complete opposite of the original. If the original is P, then the negation is ~P, which is read as "not P".