1) C is because it’s exactly double the size in both width and length while the others are only gaining one or the other. For example - the a length of the squares is 5 and the length on c is 10, the base on a is 3 long, the base on c is 6. The width on a is 1 and the width on c is 2. You can put these into fractions 5/10 3/6 1/2 but they all equal 1/2. So it’s double the size of a.
Can u show us how the location looks please?
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.
We can Notice that ∠ABC = ∠ACB
We know that if Two Angles are Equal then Angles Corresponding to those Sides are also Equal.
⇒ AB = AC
⇒ 4x + 4 = 6x - 14
⇒ 6x - 4x = 4 + 14
⇒ 2x = 18
⇒ x = 9
⇒ Length of BC = 2x + 7 = 2(9) + 7 = 18 + 7 = 25 units
The 20th visitor will be the first to get both the key chain and the bumper sticker, I found this out by finding the least common multiple, or LCM, you do this by writing out the multiples of both 4 and 10, 4 8 12 16 20..... and 10 20 30... then you find the smallest number that both 4 and 10 have in common. I hope this helps, sorry if it doesn't.
