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ale4655 [162]
2 years ago
11

A company announces that it will be giving 15% of its profits to charity. If the company profits $32,000, how much money will it

give to charity?
Mathematics
1 answer:
tankabanditka [31]2 years ago
6 0

Answer:

$4,800

Step-by-step explanation:

15% of $32,000

15% × 32000

0.15 × 32000 = $4,800

15% of $32,000 is $4,800 which will go to charity :)

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Solve for x 3(x+2)+4=5x+7
GaryK [48]
Distribute: 3x+6+4=5x+7
Simplify: 3x+10=5x+7
Subtract 3x from both sides: 10=2x + 7
Subtract 7 from both sides: 3=2x
Divide by 2: x=3/2

x=3/2
3 0
3 years ago
How does one explain how to graph a point on the coordinate grid?
Neko [114]

Answer:

Firstly you have to know the x and y points,

then you use the coordinate grid to plot the points and then draw the graph from the points that has been plotted.

Hopefully this helps :)

8 0
3 years ago
Let −→ ab= h2, −3, −1i. if the coordinates of a are (2, −1, 5), what are the coordinates of the point b?
Over [174]
The answer to that is -2 and -3
5 0
3 years ago
Which expressions are equivalint to 3x +2(x - 1) -4
Lerok [7]

Answer:

A.  3x +2(x - 1) -4 and B. 5x-6

Step-by-step explanation:

One of the answers is letter A because its the same exact expression as stated in the question. The other answer is B because of the distribution property you multiple x by 2 and multiply -1 by 2. This now gives us the expression 3x+2x-2-4. Now we add/subtract like terms and that gives the the answer 5x-6. I hope this helps.

4 0
3 years ago
Read 2 more answers
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
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