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yulyashka [42]
3 years ago
12

a restaurant offers a lunch special in which a customer can select from one of the 7 appetizers, one of the 10 entrees, and one

of the 6 desserts. How many different lunch specials are possible?
Mathematics
1 answer:
Amiraneli [1.4K]3 years ago
6 0

Answer:

420

Step-by-step explanation:

Each of the 7 appetizers can be paired with one of 10 entrees, and each entree can be paired with one of 6 desserts.  So the number of combinations is:

7 × 10 × 6 = 420

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Given: Q=7m + 3n, R= 11 - 2m, S= n + 5, and T= -m - 3n + 8 <br><br><br> simplify R - [S + T]
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R = 11-2m
S = n+5
T = -m-3n+8
---------------------------
Add up S and T to get...
S+T = (n+5)+(-m-3n+8)
S+T = n+5-m-3n+8
S+T = -m+(n-3n)+(5+8)
S+T = -m+(1n-3n)+(5+8)
S+T = -m+(-2n)+(13)
S+T = -m-2n+13
---------------------------
Subtract that result from R
R - [S + T]
11-2m - [-m-2n+13]
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(-2m+1m) + (2n) + (11-13)
(-1m) + (2n) + (-2)
-m + 2n - 2
---------------------------
---------------------------
The final answer is -m + 2n - 2
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Step-by-step explanation:

Hope this help! :D

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