Anything raised do 0 is always 1
I hope that’s correct
You have to derive for a multiplication in both terms:
=e^x+xe^x-(e^x-1 + (x-2)e^x-1) now apply distributive property in the last term:
=e^x+xe^x+e^x-1-xe^x-1 now replace each x by 0 (x=0)
=1 + 0 + e^-1 + 0 = 1+ e^-1 = 1.3679
Answer:
The pattern is this: I create a function p(x) such that
p(1)=1
p(2)=1
p(3)=3
p(4)=4
p(5)=6
p(6)=7
p(7)=9
Therefore, trivially evaluating at x=8 gives:
p(8)= 420+(cos(15))^3 -(arccsc(0.304))^(e^56) + zeta(2)
Ok, I know this isn’t what you were looking for. Be careful, you must specify what type of pattern is needed, because the above satisfies the given constraints.
Step-by-step explanation:
Answer:
y = 2x+10
Step-by-step explanation:
y is the cost (which we don't know)
2x is the $2 per card with x representing the number of cards (which we don't know)
And of course 10 is the cover charge.
