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4 years ago

Help with just 3 please.

Mathematics

1 answer:

svet-max [94.6K]4 years ago

4 0

All you have to do is:

Make all of your fractions to common denominator.
Then, you add 8 3/10-1 2/5- 2 1/3+ 4 3/5
And when you solve it, you get 9 5/30 and when you simplify it,
You get 9 1/6 yards of ribbon

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B) $\left\{\begin{matrix}a(1)=1 & \\ a(n)=a(n-1)*\frac{1}{4} &\:for\:n=1,2,3,4,... \end{matrix}\right.$

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Step-by-step explanation:

1) Incomplete question. So completing the several terms: $\left \{a_{n}\right \}_{n=1}^{\infty}=\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},... \right \}$

We can realize this a Geometric sequence, with the ratio equal to:

$q=\frac{1}{4}$

A) To find the next two terms of this sequence, simply follow multiplying the 5th term by the ratio (q):

$\frac{1}{256}*\mathbf{\frac{1}{4}}=\frac{1}{1024}\\\\\frac{1}{1024}*\mathbf{\frac{1}{4}}=\frac{1}{4096}\\\\\left \{ 1,\frac{1}{4},\frac{1}{16},\frac{1}{64},\frac{1}{256},\mathbf{\frac{1}{1024},\frac{1}{4096}}\right \}$

B) To find a recurrence a relation, is to write it a function based on the last value. So that, the function relates to the last value.

$\left\{\begin{matrix}a(1)=1 & \\ a(n)=a(n-1)*\frac{1}{4} &\:for\:n=1,2,3,4,... \end{matrix}\right.$

C) The explicit formula, is one valid for any value since we have the first one to find any term of the Geometric Sequence, therefore:

$\\a_{n}=nq^{n-1} \:for\:n=1,2,3,4,...$

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Assoli18 [71]

True
Hope this helps
Have a wonderful day or night

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