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sergeinik [125]
2 years ago
15

Solve for k: -3k = m

Mathematics
2 answers:
Tema [17]2 years ago
8 0

Answer: k = -m/3

Step-by-step explanation: Whenever you solve for k, you isolate it:

-3k = m

k = -m/3

k is now by itself.

TiliK225 [7]2 years ago
7 0

Answer:

k= -m/3

Step-by-step explanation:

thx..

......ok...

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Marissa has 2.5 hours of swim practice per day. She practices 4 days per week. The swimming season is 14 weeks long.
AveGali [126]

Answer: 140

Step-by-step explanation: 2.5 x 4 = 10

10 x 14 = 140

you multiply the amount of hours she swims for 4 days and then multiply those hours by 14 because the swimming season is 14 weeks long.

7 0
3 years ago
(x,5), (-3,6); slope = -1
anyanavicka [17]

Answer:

what exactly is the question to the problem ? it looks like it's already solved ? are you suppose to find something or ?

5 0
3 years ago
Simplify:<br> (4xy^-6)^-3 <br><br><img src="https://tex.z-dn.net/?f=%284xy%5E%7B-6%7D%20%29%5E%7B-3%7D" id="TexFormula1" title="
Zepler [3.9K]

Answer:

(4^-3)(x^-3) y^(18)

or

y^18/ (64 x^3)

Step-by-step explanation:

(4xy^-6)^-3

Distribute the exponent to all terms in the parentheses

(4^-3)(x^-3) (y^-6)^-3

a^ b^c = a^ (b*c)

(4^-3)(x^-3) y^(-6*-3)

(4^-3)(x^-3) y^(18)

If we do not want negative exponents

a ^ -b = 1/ a^b

y^18/( 4^3 x^3)

y^18/ (64 x^3)

4 0
4 years ago
Which expressions are equivalent to 64 Superscript 1? Check all that apply.
LenaWriter [7]

Answer:

  • 4³
  • 8²
  • 2⁶

Step-by-step explanation:

The factoring of 64 is ...

  64 = 2·2·2·2·2·2

  = (2·2)·(2·2)·(2·2) = 4·4·4 = 4³

  = (2·2·2)·(2·2·2) = 8·8 = 8²

  = 2·2·2·2·2·2 = 2⁶

3 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
4 years ago
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