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:
4.969696 hours
Step-by-step explanation:
The distance traveled is shown through the following equations
Distance = velocity * time
This means that
4100=velocity*4
This means that ariplane 1 travles at 1025 km/hr
We can then subtract 200 from this to find that airplane two travels at a speed of 825 km/hr
Now we need to find how long it takes for that plane to travel 4100
So using the same equation
4100=825*time
Divison will tell us that the answer is around 4.9696 hours which is pretty much 5
Answer:
9,450
Step-by-step explanation:
10 cant go into the others because they end with a 5
