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3 years ago

If six more than the product of a number and -2 is greater than 10, which of the following could be that number

1 answer:

7 0

To get that number in the problem, you need to translate first the word problem into mathematical expression.

We will let x as the number.
six more than the product of a number and -2 is greater than 10 can be written as (-2x) + 6 > 10
We will now solve for x to get the number.
   (-2x) + 6 > 10
Add -6 to both sides of the equation
      -2x + 6 - 6 > 10 - 6
      -2x > 4
Divide both sides of the equation with -2
Take note that if we divide a negative number in inequality, the inequality symbol will be change from greater than to less or from less than to greater than.
   -2x / -2< 4 / -2
   x < -2

Therefore, the number that we are looking for is less than -2.

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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

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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)

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

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

f(9) = ln(9)

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

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

Find out more information about taylor series here

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hi! so i tried to solve this one for you, let me know if it is right..! have a great day

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