A recursive sequence is a sequence of numbers whose values are determined by the numbers that come before them in the sequence.
We’re given a sequence whose (n + 1)-th term f(n + 1) depends on the value of the n-th term f(n), specified by the recursive rule
f(n + 1) = -4 f(n) + 3
We’re also given the 1st term in the sequence, f(1) = 1. Using this value and the recursive rule, we can find the next term f(2). (Just replace n with 1.)
f(1 + 1) = -4 f(1) + 3
f(2) = -4 • 1 + 3
f(2) = -1
We do the same thing to find the next term f(3) :
f(2 + 1) = -4 f(2) + 3
f(3) = -4 • (-1) + 3
f(3) = 7
One more time to find the next term f(4) :
f(3 + 1) = -4 f(3) + 3
f(4) = -4 • 7 + 3
f(4) = -25
3b(3-b) is the answer
Step by step explanation : take out the most common factor which is 3b
Answer:
(-5, 2)
Step-by-step explanation:
g(x) = f(x -h) +k
is a translation of f(x) h units to the right and k units upward. Here, we seem to have h=-5 and k=2. That means the point (0, 0) has been translated 5 units to the left (to -5) and 2 units upward (to 2).
The translated location of (0, 0) is (-5, 2).
_____
We assume we're to ignore the apostrophe (') in the equation for g(x).
In order to answer the above question, you should know the general rule to solve these questions.
The general rule states that there are 2ⁿ subsets of a set with n number of elements and we can use the logarithmic function to get the required number of bits.
That is:
log₂(2ⁿ) = n number of <span>bits
</span>
a). <span>What is the minimum number of bits required to store each binary string of length 50?
</span>
Answer: In this situation, we have n = 50. Therefore, 2⁵⁰ binary strings of length 50 are there and so it would require:
log₂(2⁵⁰) <span>= 50 bits.
b). </span><span>what is the minimum number of bits required to store each number with 9 base of ten digits?
</span>
Answer: In this situation, we have n = 50. Therefore, 10⁹ numbers with 9 base ten digits are there and so it would require:
log2(109)= 29.89
<span> = 30 bits. (rounded to the nearest whole #)
c). </span><span>what is the minimum number of bits required to store each length 10 fixed-density binary string with 4 ones?
</span>
Answer: There is (10,4) length 10 fixed density binary strings with 4 ones and
so it would require:
log₂(10,4)=log₂(210) = 7.7
= 8 bits. (rounded to the nearest whole #)
No solution
because a number can not be less than and greater than the same number.
