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.
<span>Y is directly proportional to x^2. It could be represented by the expression:
y </span>α x^2
We can make it into an equality by inserting the proportionality constant, k.
y = kx^2
k would be constant for any value of y with a corresponding value of x. We solve the problem by this concept as follows:
y1/(x1)^2 = y2/(x2)^2
10/(x1)^2 = y2/(x1/2)^2
10/4 = y2
Therefore, when the value of x is halved, y is equal to 10/4.
Answer:
Each book costs $27.
Step-by-step explanation:
Subtract 254 - 11.
254 - 11 = 243
Now, divide 243 by 9.
243 ÷ 9 = 27
Hope this helped!
It will take her 62.5 min to finish her homework
Answer:
the answer to number one is (13.3,2)
the answer to number two is (2,-2)
Step-by-step explanation:
