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4 years ago

Can someone please answer this question please answer it correctly and please show work please help me I need it

Mathematics

2 answers:

astraxan [27]4 years ago

6 0

Answer:

Step-by-step explanation:

−4−(−7)

=−4−(−7)

=−4+7

Ivanshal [37]4 years ago

3 0

Answer:

Step-by-step explanation:

Because -(-7)=+7 and we start at -4, we would move 4 units left, then 7 units right.

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Lmc and gfc of 3,5 and 6

Sholpan [36]

Answer:

Lcm is 30 and gcf is 1

Step-by-step explanation:

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3 years ago

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A farmer feeds her four horses 64 bales of hay each week. How many bales of hay would the farmer need in order to feed six horse

jekas [21]

Answer:

96 bales

Step-by-step explanation:

$\frac{64}{4}$ = $\frac{x}{6}$

You must multiply 6 by 64, which is 384.

Then you divide that by 4, which is 96.

Therefore, the answer is 96.

Also, I lined up the bales in the same row and horses in the same row, so this way can help. It can also help if it's in the same column though, but I hope this helped you!

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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

3 years ago

8 over 9 + 7 over 9 bjf

valkas [14]

Answer:

That would be 0.87

Step-by-step explanation:

Is that the answer you wanted...?

6 0

3 years ago

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