Answer:
1) The solve by graphing will the preferred choice when the equation is complex to be easily solved by the other means
Example;
y = x⁵ + 4·x⁴ + 3·x³ + 2·x² + x + 3
2) Solving by substitution is suitable where we have two or more variables in two or more (equal number) of equations
2x + 6y = 16
x + y = 6
We can substitute the value of x = 6 - y, into the first equation and solve from there
3) Solving an equation be Elimination, is suitable when there are two or more equations with coefficients of the form, 2·x + 6·y = 23 and x + y = 16
Multiplying the second equation by 2 and subtracting it from the first equation as follows
2·x + 6·y - 2×(x + y) = 23 - 2 × 16
2·x - 2·x + 6·y - 2·y = 23 - 32
0 + 4·y = -9
4) An example of a linear system that can be solved by all three methods is given as follows;
2·x + 6·y = 23
x + y = 16
Step-by-step explanation:
Question-
A team math contest has $11$ teams of $6$ students each. Before the contest begins, every student high-fives every other student who is not on their team. How many high-fives take place in all?
Answer: (-4 ,4)
Step-by-step explanation:
It might just be one or zero but I’m not sure
There is a 7/8 chance there will be boys and a 1/8 chance there will be a girl
