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3 years ago

How to get the 2 multiplication numbers for 17

Mathematics

2 answers:

tatuchka [14]3 years ago

3 0

??? Do you mean 17 divided by 2 cuz that's 8.5

Darina [25.2K]3 years ago

3 0

1,17 are the only numbers that evenly go into 17

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3 years ago

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3 years ago

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2 years ago

Find an explicit rule for the nth term of the sequence.The second and fifth terms of a geometric sequence are 18 and 144, respec

vova2212 [387]

Given:

second term = 18

fifth term = 144

The nth term of a geometric sequence is:

$\begin{gathered} a_n\text{ = ar}^{n-1} \\ Where\text{ a is the first term} \\ r\text{ is the common ratio} \end{gathered}$

Hence, we have:

$\begin{gathered} \text{ar}^{2-1}\text{ = 18} \\ ar\text{ = 18} \\ \\ ar^{5-1}=\text{ 144} \\ ar^4\text{ =144} \end{gathered}$

Divide the expression for the fifth term by the expression for the second term:

$\begin{gathered} \frac{ar^4}{ar}\text{ = }\frac{144}{18} \\ r^3\text{ = }\frac{144}{18} \\ r\text{ = 2} \end{gathered}$

Substituting the value of r into any of the expression:

$\begin{gathered} ar\text{ = 18} \\ a\text{ }\times\text{ 2 = 18} \\ Divide\text{ both sides by 2} \\ \frac{2a}{2}\text{ =}\frac{18}{2} \\ a\text{ = 9} \end{gathered}$

Hence, the explicit rule for the sequence is:

$a_n\text{ = 9\lparen2\rparen}^{n-1}$

5 0

1 year ago

Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that

kherson [118]

Answer:

The answer is

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

Step-by-step explanation:

Remember that Taylor says that

$f(x) = {\displaystyle \sum\limits_{k=0}^{\infty} \frac{f^{(k)}(a) }{k!}(x-a)^k }$

For this case

$f^{(0)} (-2) = 8(-2)-3(-2)^3 = 8\\f^{(1)} (-2) = 8-3(3)(-2)^2 = -28\\f^{(2)} (-2) = -3(3)2(-2) = -36\\f^{(2)} (-2) = -3(3)2 = -18$

$f(x) = {\displaystyle 8 + \frac{-28}{1}(x+2)+\frac{-36}{2!}(x+2)^2 + \frac{-18}{3!}(x+2)^2 }$

5 0

3 years ago