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Ede4ka [16]
3 years ago
13

You put some money in a bank account one year ago. You have not done anything with it since. The bank takes out a low balance fe

e of 0. $.25 each month. After eight months, you had $28 in the account. Choose the equation the best shows the money in your bank account over time.
Mathematics
1 answer:
gulaghasi [49]3 years ago
5 0

m - ( 8 * $0.25 ) = $ 28

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Can you help me with these questions???
lord [1]
3. 28 Quarters and 5 dimes.
3 0
3 years ago
2x+3y=41 and 5x+4y=85
Sever21 [200]

Answer:

5x+4y=85-

 Slope = -2.500/2.000 = -1.250

 x-intercept = 85/5 = 17

 y-intercept = 85/4 = 21.25000

2x+3y=41-

 Slope = -1.333/2.000 = -0.667

 x-intercept = 41/2 = 20.50000

 y-intercept = 41/3 = 13.66667

5 0
3 years ago
Divide the polynomials:<br> x^2-3x+9/x-2
Kruka [31]

The value of dividing x^2-3x+9 by x-2 is x + 3

<h3>Ways of dividing polynomials</h3>

There are several ways to divide polynomial functions; some of these ways include

  • By factorization
  • By long division
  • By synthetic division
  • By using technology such as graph

<h3>How to divide the polynomials?</h3>

The expression for the polynomial division is given as:

x^2-3x+9/x-2

To divide polynomial functions, we make use of the division by factorization method

Start by expanding the numerator of the polynomial division

x^2-3x+9/x-2 = x^2 + 3x - 6x + 9/x-2

Factorize the equation

x^2-3x+9/x-2 = x(x + 3) - 2(x + 3)/x - 2

Factor out x + 3 from the numerator

x^2-3x+9/x-2 = (x- 2)(x + 3)/x - 2

Cancel out the common factors

x^2-3x+9/x-2 = x + 3

Hence, the value of dividing x^2-3x+9 by x-2 is x + 3

Read more about polynomial division at:

brainly.com/question/25289437

#SPJ1

5 0
2 years ago
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
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

\ln(x) = \ln 3 + \ln(1+y).

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
3 years ago
How do I do this problem?
Vikentia [17]
You use the distributive property and multiple everything inside of the parentheses by 2x^2 then everything in the parentheses by y to get 2x^4y-6x^3y^2+16x^2y^3. I think that’s the answer and I hope that helped!
8 0
3 years ago
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