17 11 9 and 7
Remember that the third side must be greater than 5 and less than 19.
The first term, a, is 2. The common ratio, r, is 4. Thus,
a_(n+1) = 2(4)^(n).
Check: What's the first term? Let n=1. Then we get 2(4)^1, or 8. Is that correct? No.
Try this instead:
a_(n) = a_0*4^(n-1). Is this correct? Seeking the first term (n=1), does this formula produce 2? 2*4^0 = 2*1 = 2. YES.
The desired explicit formula is a_(n) = a_0*4^(n-1), where n begins at 1.
I need a image otherwise I can't answer this question
Hello,
Let n-2,n-1,n,n+1 the foru numbers
s=n-2+n-1+n+n+1=4n-2
95<=4n-2<=105
==>97<=7n<=107
==>24.25<=n<=26.75
==>25<=n<=26
if n=25 s=98
if n=26 s=102
f(x) = -10
set the equation equal to -10 and solve for x:
-10 = -x-1
Add 1 to both sides:
-9 = -x
Divide both sides by -1:
x = 9