Hello,
y=2^(-x)
y=2^(2x)+3
==>2^(2x)+3=1/2^x
==>2^(3x)+3*2^x-1=0 (1)
Let's assume u=2^x
(1)==>u^3+3*u-1=0
which as 3 roots
u=0.322185354626 or
u = -0.161092677313 + i1.754380959784 or
u = -0.161092677313 - i1.754380959784.
Let's take the real solution
0.322185354626=2^x
==>x=ln(0.322185354626) / ln(2)
x=-1,6340371790199...
an other way is
f(x)=2^(3x)+3*2^x-1
f(-2)=1/64+3/4-1=-15/64 <0
f(-1)=1/8+1-1=1/8>0
==> there is a solution betheen -2<x<-1
I believe the answer is D.
3^2 is 9 and 4^2 is 16, and because 9/16 was the third term in the sequence, you raise both the numerator and the denominator by the number of the term minus 1.
So to find the seventh term, you raise the numerator and denominator by 7 - 1 = 6.
So the answer is 729/4096
F (x+h)-f (x)/h
<------------------------
Answer:
Number 2
Step-by-step explanation:
Go from the largest negative number then go down the numbers till 0 then go up with the rest of the number is order.