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:
1. The equation is given as
324 = 220 +135 + C
2. We have 166 As the correct answer but is not in the option
Step-by-step explanation:
We are been given the following values
No. of trophies = 3
Total Legos required for all trophies = 324
No. of Legos for largest trophy(A) = 220
No. of Legos for smallest trophy(B) = 135
No. of Legos required for final trophy(C) which is unknown
Total legos will addition of everything
Total Legos = [Number of Legos for largest trophy(A) + Number of Legos for smallest trophy(B) + Number of Legos required for final trophy(C)]
Substituting the values
324 = 220 +135 + C
Making C subject of formula
324 = 355 + C
324 - 355 = C
-31 = C
135 + 31 = 166
We have 166 As the answer
Answer:
3/2 or 1.5 or 1 1/2
Step-by-step explanation:
Y= mx +b ; x is the slope and y is the y-intercept.
So in this case, our equation will be y= 3x + 19 :)
Answer:
80
Step-by-step explanation:
because 8 is in tens place
so it´s value is 8*10=80
