Answer:if A= [2 6] prove that (A')' = A
3 7
4 9
Step-by-step explanation:
A) I would make the positive integer x and then form an equation.
x + 30 = x^2 - 12
x + 42 = x^2
0 = x^2 - x - 42 this can be factorised
(x - 7) ( x + 6) Therefore x = 7 or x = -6
Since the question asks for a positive integer the answer is 7.
B) two positive numbers x and y.
X - y = 3
x^2 + y^2 = 117
Use these simultaneous equations to figure out each number.
Rearrange the first equation
x = y + 3
Then substitute it into the second equation.
(y+3)^2 + y^2 = 117
y^2 + 6y + 9 + y^2 = 117
2y^2 + 6y - 108 = 0
then factorise this.
(2y - 12) (y + 9)
This means that y = 6 or y = -9 but it’s 6 because that’s the only positive number.
Use y to find x
x = y + 3
x = 6 + 3
x = 9
So the answers are x = 9 and y = 6.
Its A a single outlier causes the upper quartiles to move close together
im pretty sure
