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SCORPION-xisa [38]

3 years ago

10

Write the prime factorization of 10. Use exponents when appropriate and order the factors from least to greatest (for example, 2

* 2*3*5)

1 answer:

earnstyle [38]3 years ago

5 0

The prime factorization of 10 is 2*5. No exponents required.

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This problem asks for Taylor polynomials forf(x) = ln(1 +x) centered at= 0. Show Your work in an organized way.(a) Find the 4th,

stich3 [128]

Answer:

a) The 4th degree , 5th degree and sixth degree polynomials

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

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

b)The nth degree Taylor polynomial for f(x) centered at x = 0, in expanded form.

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

Step-by-step explanation:

Given the polynomial function f(x) = log(1+x) …...(1) centered at x=0

f(x) = log(1+x) ……(1)

using formula $\frac{d}{dx} logx =\frac{1}{x}$

Differentiating Equation(1) with respective to 'x' we get

$f^{l} (x) = \frac{1}{1+x} (\frac{d}{dx}(1+x)$

$f^{l} (x) = \frac{1}{1+x} (1)$ ….(2)

At x= 0

$f^{l} (0) = \frac{1}{1+0} (1)= 1$

using formula $\frac{d}{dx} x^{n-1} =nx^{n-1}$

Again Differentiating Equation(2) with respective to 'x' we get

$f^{l} (x) = \frac{-1}{(1+x)^2} (\frac{d}{dx}((1+x))$

$f^{ll} (x) = \frac{-1}{(1+x)^2} (1)$ ….(3)

At x=0

$f^{ll} (0) = \frac{-1}{(1+0)^2} (1)= -1$

Again Differentiating Equation(3) with respective to 'x' we get

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (\frac{d}{dx}((1+x))$

$f^{lll} (x) = \frac{(-1)(-2)}{(1+x)^3} (1)= \frac{(-1)^2 (2)!}{(1+x)^3}$ ….(4)

At x=0

$f^{lll} (0) = \frac{(-1)(-2)}{(1+0)^3} (1)$

$f^{lll} (0) = 2$

Again Differentiating Equation(4) with respective to 'x' we get

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (\frac{d}{dx}((1+x))$

$f^{lV} (x) = \frac{(2(-3))}{(1+x)^4} (1)= \frac{((-1)^3(3!))}{(1+x)^4}$ ....(5)

$f^{lV} (0) = \frac{(2(-3))}{(1+0)^4}$

$f^{lV} (0) = -6$

Again Differentiating Equation(5) with respective to 'x' we get

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} (\frac{d}{dx}((1+x))$

$f^{V} (x) = \frac{(2(-3)(-4))}{(1+x)^5} =\frac{(-1)^4 (4!)}{(1+x)^5}$ .....(6)

At x=0

$f^{V} (x) = 24$

Again Differentiating Equation(6) with respective to 'x' we get

$f^{V1} (x) = \frac{(2(-3)(-4)(-5))}{(1+x)^6} (\frac{d}{dx}((1+x))$

$f^{V1} (x) = \frac{(-120))}{(1+x)^6} (1) = \frac{(-1)^5 5!}{(1+x)^6}$

and so on...

The nth term is

$f^{n} (x) = = \frac{(-1)^{n-1} (n-1)!}{(1+x)^n}$

Step :-2

Taylors theorem expansion of f(x) is

$f(x) = f(a) + \frac{x}{1!} f^{l}(x) +\frac{(x-a)^2}{2!}f^{ll}(x)+\frac{(x-a)^3}{3!}f^{lll}(x)+\frac{(x-a)^4}{4!}f^{lV}(x)+\frac{(x-a)^5}{5!}f^{V}(x)+\frac{(x-a)^6}{6!}f^{VI}(x)+...….. \frac{(x-a)^n}{n!}f^{n}(x)$

At x=a =0

$f(x) = f(0) + \frac{x}{1!} f^{l}(0) +\frac{(x)^2}{2!}f^{ll}(0)+\frac{(x)^3}{3!}f^{lll}(0)+\frac{(x)^4}{4!}f^{lV}(0)+\frac{(x)^5}{5!}f^{V}(0)+\frac{(x)^6}{6!}f^{VI}(0)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

Substitute all values , we get

$f(x) = f(0) + \frac{x}{1!} (1) +\frac{(x)^2}{2!}(-1)+\frac{(x)^3}{3!}(2)+\frac{(x)^4}{4!}(-6)+\frac{(x)^5}{5!}(24)+\frac{(x)^6}{6!}(-120)+...….. \frac{(x-0)^n}{n!}f^{n}(0)$

On simplification we get

$log(1+x) = x - \frac{x^2}{2} +\frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6}+\\.. (-1)^{n-1}\frac{x^n}{n} +..$

4 0

4 years ago

Suppose a study estimated that 63% of the residents of a community (with an

Anvisha [2.4K]

Answer:

Answer choice A(57%)

Step-by-step explanation:

The margin of error is 7%. This means the 63% could either 7% too high or 7% too low. Answer choice A is the only answer choice that is within the range of percents that it could possibly be(between 56% and 70%). Answer choice B is 10% off, 3% out of the range. Answer choice C is 25% off, 18% out of the range. Answer choice D is 32% off, 24% out of the range. Hope this helps. :)

3 0

4 years ago

(PLEASE HELPPP!!!)

lana [24]

I found the value for number 2 and it took some time to like calculate but here it is
$15,250
1.)$14,106.25
2.)$13,048.2812
3.)$12,069.6601
4.)$11,164.4356
5.)$10,327.1029
6.)$9,552.57018
7.)$8,836.12742
8.)$8,173.42
The numbers are the years but the value of the car would be number 8

7 0

3 years ago

How do i Solve this equation<br> √x+5-2=0

grigory [225]

Answer:

9

Step-by-step explanation:

√x+5-2=0

√x+3=0

√x= -3

x=9 [(-3)*(-3) = 9]

8 0

2 years ago

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Your gym membership costs $33 per monthafter an initial membership fee. you paid a total of $228 after 6 months

Maru [420]

1. let x be the number of months paid. 2. let 33x be the dollar value of the cost paid after x months (no initial fee yet) 3. let 33*5=165 be the cost paid on monthly basis for 5 months . But the total cost is $225 dollers after 5 months. the implies that the membership fee is 225 -165=dollers. 4. let TC be the total cost. 5. then the total cost TC=33x+60. 6. after 10 months , total cost TC= 33*10+60=330+60=390dollers.

7 0

4 years ago

Read 2 more answers