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:
3.16 ≈ 3.2
Step-by-step explanation:
Put the numbers in the distance formula and do the arithmetic.
d = √((x2 -x1)² +(y2 -y1)²)
d = √((2-(-1))² +(3-2)²) = √(3² +1²) = √10
d ≈ 3.16 ≈ 3.2
The distance rounded to hundredths is 3.16. Rounded to tenths, it is 3.2.
//// For G(x) = -1/4x - 1////
G(-5) = -1/4(-5) - 1 = 0.25;
G(-2) = -1/4(-2) - 1 = -0.25;
G(-1) = -1/4(-1) - 1 = -0.75;
///////////////////////////////////
////For G(x) = (x-1)^2 - 3////
G(-1) = ((-1) -1)^2 - 3 = 1;
G(-2) = ((-2) - 1)^2 - 3 = 6;
G(-5) = ((-5) - 1)^2 - 3 = 33;
//////////////////////////////////////////
////For G(x) = 1/4x + 2////////
G(-5) = 1/4(-5) + 2 = 0.75;
G(-2) = 1/4(-2) + 2 = 1.5;
G(-1) = 1/4(-1) + 2 = 1.75;