Answer:
C = 100 + 40n
The independent variable is n (number of months) and the dependent variable is C ( total cost)
Step-by-step explanation:
Let the total cost of the fitness center = C
Given fixed cost = $100
given cost per month = $40
let number of month = n
The total cost of the fitness center in a given 'n' month is calculated as;
C = 100 + 40n
From the above equation;
the independent variable is n (number of months) and
the dependent variable is C ( total cost)
Answer: The price for the adult tickets is $12 and the price for the child ticket is $5.
Step-by-step explanation:
Let the price for the adult ticket be x
Let the price for the child ticket be y
According to question, On Friday ,

and on next day,

Now, we will use "Substitution Method" to solve the system of equations :

so, we will put the value of x in the first equation i.e.

Now, put the value of y in the equation that is given by

Hence, the price for the adult tickets is $12 and the price for the child ticket is $5.
Answer:
The value of the given expression is -878.08.
Step-by-step explanation:
Here, the given question is INCOMPLETE.
The complete question is:
Evaluate the expression for a = 4, b = -5, and c = -7. 
Here, substitute the vale of a = 4, b = -5 and c = -7, we get:

Hence, the value of the given expression is -878.08.
By Direct Proof :
<span><span>1.(∼H(x)∨∼S(x))→(P(x)∨L(x))</span><span>1.(∼H(x)∨∼S(x))→(P(x)∨L(x))</span></span>
<span><span>2.P(x)→E(x)</span><span>2.P(x)→E(x)</span></span>
<span><span>3.∼E(x)</span><span>3.∼E(x)</span></span>
<span><span>−−−−−−−−−−−−−−−−−−−−−</span><span>−−−−−−−−−−−−−−−−−−−−−</span></span>
<span><span>4.H(x)</span><span>4.H(x)</span></span>
<span><span>5.P(x)→E(x)≡∼P(x)∨E(x)</span><span>5.P(x)→E(x)≡∼P(x)∨E(x)</span></span> by Material Implication
<span><span>6.∼P(x)</span><span>6.∼P(x)</span></span> , #5 and #3 by Disjunctive Syllogism
<span><span>7.∼P(x)∨∼L(x)</span><span>7.∼P(x)∨∼L(x)</span></span> , #6 by Addition ( I just add <span>∼∼</span>L(x))
Since #7 is logically equivalent to <span><span>∼(P(x)∨L(x))</span><span>∼(P(x)∨L(x))</span></span> by De Morgan's Law,
<span><span>8.∼(∼H(x)∨∼S(x))</span><span>8.∼(∼H(x)∨∼S(x))</span></span> , #1 and #7 by Modus Tollens.
Distributing the <span>∼∼</span>, we'll have,
<span><span>9.H(x)∧S(x)</span><span>9.H(x)∧S(x)</span></span> by De Morgan's and Double Negation
<span><span>10.H(x)</span><span>10.H(x)</span></span> by Simplification <span>■</span>