Answer:
Let's define the cost of the cheaper game as X, and the cost of the pricer game as Y.
The total cost of both games is:
X + Y
We know that both games cost just above AED 80
Then:
X + Y > AED 80
From this, we want to prove that at least one of the games costed more than AED 40.
Now let's play with the possible prices of X, there are two possible cases:
X is larger than AED 40
X is equal to or smaller than AED 40.
If X is more than AED 40, then we have a game that costed more than AED 40.
If X is less than or equal to AED 40, then:
X ≥ AED 40
Now let's take the maximum value of X in this scenario, this is:
X = AED 40
Replacing this in the first inequality, we get:
X + Y > AED 80
Replacing the value of X we get:
AED 40 + Y > AED 80
Y > AED 80 - AED 40
Y > AED 40
So when X is equal or smaller than AED 40, the value of Y is larger than AED 40.
So we proven that in all the possible cases, at least one of the two games costs more than AED 40.
Answer:
8 + 1.5x + 0.75y = pizza_cost
Step-by-step explanation:
It is given that every 10-inch cheese pizza costs ($8), meaning that this will be the constant.
Let (x) represent the number of inches added to the pizza. Since each extra additional inch costs ($1.5) thus one adds (1.5x) to the equation.
Let (y) represent the number of toppings. Every extra topping costs ($0.75) therefore one has to add (0.75y) to the equation.
Putting all of the elements together one gets the equation,
8 + 1.5x + 0.75y
Close enough, it would be 36+6x.
Answer:
B:5x
Step-by-step explanation:
It would be B because if you add all the sides together you would get 5x and 5x is B
