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4 years ago

Tom needs to buy four tickets at 16 dollars each with a service fee of 2.50 . Write an expression to solve for n

Mathematics

1 answer:

Lorico [155]4 years ago

7 0

Answer:

16*4+2.50

Step-by-step explanation:

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Find the length of the third side.if necessary, round to the nearest tenth 24 18

VashaNatasha [74]

Answer:

Step-by-step explanation:

\text{Legs: 18, 24}\hspace{20px}\text{Hypotenuse: ?}

Legs: 18, 24Hypotenuse: ?

a^{2}+b^{2}=

\,\,c^{2}

The Pythagorean Theorem

18^{2}+24^{2}=

+24

\,\,c^{2}

Plug in. c is the hypotenuse.

324+576=

\,\,c^{2}

Square the numbers.

900=

\,\,c^{2}

Add.

\sqrt{900}=

900

\,\,\sqrt{c^{2}}

Square root both sides.

30=

\,\,c

The hypotenuse.

7 0

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

What is the length of segment AC

lakkis [162]

To find the length of segment AC, we must find the total rise and total run between the two points.

Point C is located at (-5, 5). Point A is located at (3,-1). To find the rise, subtract the y-value of A from the y-value of C:

$5 - (-1) = 6$

The rise of this segment is 6.

To find the run, subtract the x-value of A from the x-value of C:

$3 - (-5) = 8$

The run of this segment is 8.

We can use the Pythagorean Theorem to find the length of this segment. The theorem uses the following formula:

$a^{2} + b^{2} = c^{2}$

The segment represents the hypotenuse, and the rise and run represent the legs of this segment. We know that the two legs' lengths are 6 and 8, so plug them into the equation:

$6^{2} + 8^{2} = c^{2}$
$36 + 64 = c^{2}$
$100 = c^{2}$

Square root both sides to get c by itself:

$\sqrt{100} = 10$
$c = 10$

The length of segment AC is 10.

6 0

3 years ago

Read 2 more answers

Marge is twice as old as Consuelo. The sum of their ages seven years ago was 13. How old are they now?

Trava [24]

Merge is 18.
Conseulo is 9

7 0

3 years ago

Read 2 more answers

A company that relies on Internet-based advertising linked to key age demographics wants to understand the relationship between

MA_775_DIABLO [31]

Answer:

a)D. Since the company wants to predict revenue from advertising expenditure, the explanatory variable is advertising expenditure.

b)B. Since the company wants to predict revenue from advertising expenditure, the response variable is revenue

c)C. Since advertising expenditure is the explanatory variable, it should be plotted on the x-axis.

Step-by-step explanation:

a) The explanatory variable is the independent variable (the difference between explanatory an indepent is that independent means that it doesn´t depend of any variable at all and in the explanatory variable we are not 100% sure)

In this case, the company wants to know if the money they spend in advertising is related to the company's revenue. So , in this case the explanatory variablle is the money they spend in advetirsing while the dependent variable is the revenue ( that will increase or decrease depending on the money spent)

b) The response variable ist he dependent variable so like it was explained in a) is revenue.

c) In a cartesian graphic on the x axis we always draw the independent variable, so in this case we will draw the money spent in advertising.

7 0

3 years ago