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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Since there is nothing attached for me to help you I looked for a website that can find the answer with an explanation. It is called quickmath it’s like a math calculator . Hopefully that helps you .
Answer:
D) 14 seconds
Step-by-step explanation:
First we will plug 500 in for y:
500 = -4.9t² + 120t
We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:
500-500 = -4.9t² + 120t - 500
0 = -4.9t²+120t-500
Our values for a, b and c are:
a = -4.9; b = 120; c = -500
We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.
The quadratic formula is:

Using our values for a, b and c,

The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is
19-5 = 14 seconds.
Answer:
48,58
Step-by-step explanation:
Let The speed of the west bound train be x
Let the east bound train be x+10
The speeds can be calculated as follows
2(2x+10)= 232
4x + 40= 232
4x= 232-40
4x= 192
x= 192/4
= 48
Hence the speedof the west bound train is 48
The speed of the east bound train is 48+10
= 58