This is a combination problem.
Given:
12 students
3 groups consisting of 4 students.
Mark can't be in the first group.
The combination formula that I used is: n! / r!(n-r)!
where: n = number of choices ; r = number of people to be chosen.
This is the formula I used because the order is not important and repetition is not allowed.
Since Mark can't be considered in the first group, the value of n would be 11 instead of 12. value of r is 4.
numerator: n! = 11! = 39,916,800
denominator: r!(n-r)! = 4!(11-4)! = 4!*7! = 120,960
Combination = 39,916,800 / 120,960 = 330
There are 330 ways that the instructor can choose 4 students for the first group
Answer:
dhkmbhuigago katangina mo
So you need to come up with a perfect square that works for the x coefficients.
like.. (2x + 2)^2
(2x+2)(2x+2) = 4x^2 + 8x + 4
Compare this to the equation given. Our perfect square has +4 instead of +23. The difference is: 23 - 4 = 19
I'm going to assume the given equation equals zero..
So, If we add subtract 19 from both sides of the equation we get the perfect square.
4x^2 + 8x + 23 - 19 = 0 - 19
4x^2 + 8x + 4 = - 19
complete the square and move 19 over..
(2x+2)^2 + 19 = 0
factor the 2 out becomes 2^2 = 4
ANSWER: 4(x+1)^2 + 19 = 0
for a short cut, the standard equation
ax^2 + bx + c = 0 becomes a(x - h)^2 + k = 0
Where "a, b, c" are the same and ..
h = -b/(2a)
k = c - b^2/(4a)
Vertex = (h, k)
this will be a minimum point when "a" is positive upward facing parabola and a maximum point when "a" is negative downward facing parabola.
The test That holds true for this inequality is given as 1/4 and 1
<h3>How to solve for the inequality</h3>
3/2 y - 2x > 1
The goal is to make y the subject
then
3/2 y > 2x + 1
We have to divide through the equation by 3/2
Such that y > 4/3 x + 2/3
Read more on inequality here: brainly.com/question/25275758
#SPJ1
