Answer:
a(n)=1.15[a(n-1)]
Step-by-step explanation:
we know that
Let
a0 -----> the length of the original copy
<em>The first copy is equal to</em>
a1=1.15(a0)
<em>The second copy is </em>
a2=1.15[1.15(a0)] or a2=1.15[a1]
<em>The third copy is</em>
a3=1.15{1.15[1.15(a0)]} or a3=1.15[a2]
therefore
A recursive formula will be
a(n)=1.15[a(n-1)]
<span>5t=3b+660....and ... 2t+5b=450
from the first we can see that t=(3b+660)/5 and using this value in the second...
(6b+1320)/5+5b=450 making all have common denominator..
6b+1320+25b=2250
31b=930
b=30...and since 2t+5b=450
t=150</span>
B=3; C=16; and E=4096.
Using the information we have gives us
We know that:
We know that 2³ = 8, so B = 3.
To "undo" a square root, we square the number; 4²=16, so C = 16.
To "undo" a fourth root, we raise the answer to the fourth power; 8⁴ = 4096, so E = 4096.
219+159≤x≤369+309
The answer can then be simplified to
378≤x≤678
