Let's let the weight of a large box be L, and the weight of a small box be S.
We know that 5 large boxes and 3 small boxes is 120kg, so:
5L + 3S = 120
We also know that 7 large boxes and 9 small boxes is 234kg, so:
7L + 9S = 234
You can multiply the first equation by 3 to get:
15L + 9S = 360
See how now both equations have 9S? We can now subtract one from the other:
(15L+9S) - (7L+9S) = 360-234
8L = 126
L = 15.75
Now sub this value back into an equation:
(5x15.75) + 3S = 120
3S = 41.25
S = 13.75
Double check these values
(7x15.75) + (9x13.75)
= 110.25 + 123.75
=234, which is consistent with above.
So a large box is 15.75kg, and a small box is 13.75kg.
Hope this helped
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:
6 minutes
Step-by-step explanation:
60 * 10% =
60 * 0.1 =
6
Assuming n is the number of signers.
7 is 10% of y.
7 = 0.1y
y = 70
n is 80% of y
n = 0.8y
n = 0.8(70)
n = 56
56 signers
You have in total 42 hours so 2 of those hours are double time so you do 40•11 and then add 22•2 (22 is the double time) so you’re answer is $484
