The number of different dinner combinations for each student that are possible is 24
<h3>How many different dinner combinations for each student are possible?</h3>
The given parameters are:
Entree choices = 3
Side dish = 4
Beverage = 2
The number of different dinner combinations for each student that are possible is
Combination = Entree * Side dish * Beverage
This gives
Combination = 3 * 4 * 2
Evaluate
Combination = 24
Hence, the number of different dinner combinations for each student that are possible is 24
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Answer:
x=9
Step-by-step explanation:
Original equation: 16x-8-14x=10
Combine like terms: 2x-8=10
Add 8 to both sides: 2x=18
Divide by 2 on both sides: x=9
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Answer:
16
Step-by-step explanation:
Since line CE is equal to 32 and line DB is the bisector (cut into the middle) of line CE then line CF would be half of line CE. Hence the reason that line CF is equal to 16
Answer:
1.98
Step-by-step explanation:
Weight on earth/ 9.81 m/s^2 * 1.622m/s^2
If C is between A and B, then, we can write the equation
Substitute the value of AB and the expressions for AC and CB in terms of x.
We can now solve for x.
Dividing by 5.
