Step 1
Given; The equations in the elimination method are written in the standard form as ____
Step 2
Simultaneous linear equations can be solved using the elimination method. First of all, make sure that the equations are written in the standard form either Ax+By=C or Ax+By+C=0. In this method, we multiply both the equations with a non-zero number to make the coefficients of any one variable equal.
Thus the answer is; The equations are written in standard form as seen below
I'm going to have to go with choice "c". 10x - 1 with the remainder of 5
I will explain you and pair two of the equations as an example to you. Then, you must pair the others.
1) Two circles are concentric if they have the same center and different radii.
2) The equation of a circle with center xc, yc, and radius r is:
(x - xc)^2 + (y - yc)^2 = r^2.
So, if you have that equation you can inmediately tell the coordinates of the center and the radius of the circle.
3) You can transform the equations given in your picture to the form (x -xc)^2 + (y -yc)^2 = r2 by completing squares.
Example:
Equation: 3x^2 + 3y^2 + 12x - 6y - 21 = 0
rearrange: 3x^2 + 12x + 3y^2 - 6y = 21
extract common factor 3: 3 (x^2 + 4x) + 3(y^2 -2y) = 3*7
=> (x^2 + 4x) + (y^2 - 2y) = 7
complete squares: (x + 2)^2 - 4 + (y - 1)^2 - 1 = 7
=> (x + 2)^2 + (y - 1)^2 = 12 => center = (-2,1), r = √12.
equation: 4x^2 + 4y^2 + 16x - 8y - 308 = 0
rearrange: 4x^2 + 16x + 4y^2 - 8y = 308
common factor 4: 4 (x^2 + 4x) + 4(y^2 -8y) = 4*77
=> (x^2 + 4x) + (y^2 - 2y) = 77
complete squares: (x + 2)^2 - 4 + (y - 1)^2 - 1 = 77
=> (x + 2)^2 + (y - 1)^2 = 82 => center = (-2,1), r = √82
Therefore, you conclude that these two circumferences have the same center and differet r, so they are concentric.
Answer:
Hello Everybody Love Pizza
Step-by-step explanation:
There are 14 chairs and 8 people to be seated. But among the 8. three will be seated together:
So 5 people and (3) could be considered as 6 entities:
Since the order matters, we have to use permutation:
¹⁴P₆ = (14!)/(14-6)! = 2,162,160, But the family composed of 3 people can permute among them in 3! ways or 6 ways. So the total number of permutation will be ¹⁴P₆ x 3!
2,162,160 x 6 = 12,972,960 ways.
Another way to solve this problem is as follow:
5 + (3) people are considered (for the time being) as 6 entities:
The 1st has a choice among 14 ways
The 2nd has a choice among 13 ways
The 3rd has a choice among 12 ways
The 4th has a choice among 11 ways
The 5th has a choice among 10 ways
The 6th has a choice among 9ways
So far there are 14x13x12x11x10x9 = 2,162,160 ways
But the 3 (that formed one group) could seat among themselves in 3!
or 6 ways:
Total number of permutation = 2,162,160 x 6 = 12,972,960
