Interesting question. Good to know for computer science.
Suppose you have a function like
an = 3x - 2 Try the first couple
a1 = 3(1) - 2
a1 = 3 - 2
a1 = 1
a2 = 3(2) - 2
a2 = 6 - 2
a2 = 4 So each term differs by 3
a2 - a1 = 3
an = a_(n - 1) + 3
a3 = a2 + 3
a3 = 4 + 3
a3 = 7
a4 = a3 + 3
a4 = 7 + 3
a4 = 10
a5 = a4+ 3
a5 = 10 + 3
a5 = 13
I'll do one more and then check it.
a6 = a5 + 3
a6 = 13 + 3
a6 = 16
a6 = 3x -2
a6 = 3*6 - 2
a6 = 18 - 2
a6 = 16 which checks.
So the general formula is
an = a_(n - 1) * k if you were multiplying or
an = a_(n - 1) + k if you were adding. The key thing is that you are working with the previous term.
Answer:
first
area of rectangle = l*b
= (3x-y+4) ( X+7)
= 3x²-xy + 4x+21x-7y+28
= 3x²-xy +25x-7y +28
Area of square = l²
= (x-2)²= x²- 4x +4
area of shaded region=( 3x²-xy +25x-7y +28) -(x²- 4x +4)
=2x²-xy +29x-7y+24
Step-by-step explanation:
Total possible outcomes are= 6×6=36
For example, if you roll a 1 on the blue dice, you cannot roll another 1 (since 1 is not bigger than 1).
Rolling a 2, you'd have 4 possible choices left (you cannot roll a 1 or 2 since they are less than 2).
(1/6)<-- picking a 1/2/3/4/5/6 out of everything;
(1/6)(5/6)=5/36 <-- (5/6) possibilities of picking a number bigger than 1
(1/6)(4/6)=4/36 <-- (4/6) possibilities of picking a number bigger than 2
(1/6)(3/6)=3/36
(1/6)(2/6)=2/36
(1/6)(1/6)=1/36
(1/6)(0)=0 <-- Impossible to get a number bigger than 6 in a normal dice
Add all of them together;
(5+4+3+2+1+0)/36=15/36
Therefore, there are 15/36 possibilities of the red die showing a number larger than that of the blue die.
Hope I helped :)