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Flauer [41]
3 years ago
15

A number m plus 4 is greater than it equal to 7

Mathematics
1 answer:
larisa [96]3 years ago
4 0
M could be anything 3 or over
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Solve the equation.5x +2(11-x)=-5
Vladimir79 [104]

Step-by-step explanation:

5x + 2(11 - x) = - 5

5x + 22 - 2x = - 5

3x = - 5 - 22

3x = - 27

x = -27 / 3

x = 9

Therefore x = 9 .

3 0
3 years ago
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Adrian has a square bedroom. The area of the room is 169 ft². How would you find the length of one side of his room?
myrzilka [38]
To find the length of one sides of the square, you must find the square root of the area. Therefore, your answer would be 13.
4 0
3 years ago
Hello, help me please))
Fed [463]

Answer:

  • 6
  • 9
  • 1/8
  • -10/11

Step-by-step explanation:

You can rewrite the given log expressions to express them in terms of ln(a), ln(b), and ln(c). Then substituting the given values will produce the value of the expression.

Or, you can define the variables 'a', 'b', and 'c' and let your calculator compute these directly.

__

<h3>1.</h3>

  \ln\left(\dfrac{a^4}{b^4c^{-2}}\right)=4\ln(a)-(4\ln(b)-2\ln(c))=4\cdot2-4\cdot3+2\cdot5=\boxed{6}

<h3>2.</h3>

  \ln\left(\sqrt{b^{-2}c^4a^2}\right)=\dfrac{1}{2}\left(-2\ln(b)+4\ln(c)+2\ln(a)\right)=\ln(a)-\ln(b)+2\ln(c)\\\\=2-3+2\cdot5=\boxed{9}

<h3>3.</h3>

  \dfrac{\ln(a^1b^2)}{\ln(bc)^2}=\dfrac{\ln(a)+2\ln(b)}{(\ln(b)+\ln(c))^2}=\dfrac{2+2\cdot3}{(3+5)^2}=\dfrac{8}{64}=\boxed{\dfrac{1}{8}}

<h3>4.</h3>

  \ln(c^{-2})\left(\ln\dfrac{a}{b^{-3}}\right)^{-1}=\dfrac{-2\ln(c)}{\ln(a)-(-3)\ln(b)}=\dfrac{-2(5)}{2+3\cdot3}=\boxed{-\dfrac{10}{11}}

_____

The applicable rules of logarithms are ...

  • ln(ab) = ln(a) +ln(b)
  • ln(a/b) = ln(a) -ln(b)
  • ln(a^b) = b·ln(a)
  • ln(a) = b  ⇔  a = e^b

Of course, a square root is the same as a 1/2 power.

7 0
2 years ago
Supplementary angles
ad-work [718]

If u need defination of supplementary angle then it is like this,

THOSE ANGLE WHICH IS EXACTLY 189 DEGREE IS CALLED SUPPLEMENTARY ANGLE.

3 0
3 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
4 years ago
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