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Dahasolnce [82]
3 years ago
9

The quotient of the sum of 2t and 2 and twice the cube of s

Mathematics
1 answer:
Vladimir79 [104]3 years ago
8 0

The equation according to the given verbal sentence is \frac{2 t+2}{2 s^{3}} (after solving this \frac{t+1}{s^{3}} ).

<u>Step-by-step explanation:</u>

Rewrite the given verbal sentence and so it will be easier to translate.

Given: The quotient of the sum of 2 t and 2 and twice the cube of s

sum of 2 t and 2 can be written as

       2 t + 2

twice the cube of s

      2 s^{3}

Now, combine the above according to the given verbal. The quotient of the sum of 2 t and 2 and twice the cube of s  means,

             2 t + 2  divided by 2 s^{3}

Now, the equation would be \frac{2 t+2}{2 s^{3}}

When solving this equation, we get

         \frac{2(t+1)}{2 s^{3}}=\frac{t+1}{s^{3}}

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Dahasolnce [82]

Step-by-step explanation:

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Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }

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Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }

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Following the pattern, we can see that for f^{n}(x),

f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} }  \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}

This applies for n ≥ 1, Expressing f(x) in summation, we have

\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

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