The answer will be 21u^6v^8
Sam has 4 dogs AND His friend Erick hace him 4 More??
<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
Answer:
the burgers be 3 and fries be 2
Step-by-step explanation:
The computation is shown below:
Let us assume burgers be x
And, the fries be y
Now according to the questiojn
1.25x + 0.50y = $4.75
1.50x + 0.99y = $6.48
Now multiply by 1.2 in the first equation
1.50x + 0.6y = $5.70
1.50x + 0.99y = $6.48
-0.39y = -0.78
y = 2
Now put the value of y in any of the above equation
1.25x + 0.50(2) = $4.75
x = 3
Hence, the burgers be 3 and fries be 2
11.4 should be the answer
if you mean tens then it is 10
