Answer:
procedure always produces 6
Step-by-step explanation:
Let 'n' be the unknown number
Add 4 to the number : n+4
multiply the sum by 3.
multiply the sum n+4 by 3
Now subtract 6, so we subtract 6 from 3n+12
finally decrease the difference by the tripe of the original number
triple of original number is 3n
so the procedure always produces 6
Answer:
Shift 2 unit left
Flip the graph about y-axis
Stretch horizontally by factor 2
Shift vertically up by 2 units
Step-by-step explanation:
Given:
Parent function: 
Transformation function: 
Take -2 common from transform function f(x)
![f(x)=\log[-2(x+2)]+2](https://tex.z-dn.net/?f=f%28x%29%3D%5Clog%5B-2%28x%2B2%29%5D%2B2)
Now we see the step-by-step translation
Shift 2 unit left ( x → x+2 )
Flip the graph about y-axis ( (x+2) → - (x+2) )
![f(x)=\log[-(x+2)]](https://tex.z-dn.net/?f=f%28x%29%3D%5Clog%5B-%28x%2B2%29%5D)
Stretch horizontally by factor 2 [ -x(x+2) → -2(x+2) ]
![f(x)=\log[-2(x+2)]](https://tex.z-dn.net/?f=f%28x%29%3D%5Clog%5B-2%28x%2B2%29%5D)
Shift vertically up by 2 units [ f(x) → f(x) + 2 ]
Simplify the function:
Hence, Using four step of transformation to get new function
Answer:
Step-by-step explanation:
1 ) Given that
For a non homogeneous part 
, we assume the particular solution is
2 ) Given that
For a non homogeneous part 
, we assume the particular solution is
3 ) Given that
y′′ + 4y′ + 20y = −3sin(2x)
For a non homogeneous part −3sin(2x) , we assume the particular solution is
4 ) Given that
y′′ − 2y′ − 15y = 3xcos(2x)
For a non homogeneous part 3xcos(2x) , we assume the particular solution is
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have:
. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.
Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold
A.
ABA and AB
A look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.
So in this case the formula would be
. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!
Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.
Answer:
$ 24.30
Step-by-step explanation:
1. $32.40 multiplied by 15 which would equal 486
2. Then you would do 486 divided by 20 which would equal $24.3
3. Then you just add the zero the the end
