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bazaltina [42]
3 years ago
8

When decreases in one variable are accompanied by decreases in another variable, the variables are described as?

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0
Directly proportional or positively correlated
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Alvin is 9 years younger than Elga. The sum of their ages is 67. What is Elga's age?
ASHA 777 [7]

Answer:

Elga is 38

Step-by-step explanation:

Alvin is 29

38 + 29 = 67

6 0
2 years ago
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Simplify (4x-6)-(3x+6)
DaniilM [7]

Answer:

x-12

Step-by-step explanation:

First open the brackets

=(4x-6)-(3x+6)\\\\\\=4x-6-3x-6

Then collect the like terms

4x-3x-6-6

Solve the like terms

4x-3x=x\\\\\\-6-6=-12\\\\\\=x-12

The simplified form is

x-12

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
3 years ago
The figure is made up of a hemisphere and a cylinder.
Goryan [66]
Data: (Cylinder)
h (height) = 8 cm
r (radius) = 5 cm
Adopting: \pi \approx 3.14
V (volume) = ?

Solving:(<span>Cylinder volume)
</span>V = h* \pi *r^2
V = 8*3.14*5^2
V = 8*3.14*25
\boxed{ V_{cylinder}  = 628\:cm^3}

<span>Note: Now, let's find the volume of a hemisphere.
</span>
Data: (hemisphere volume)
V (volume) = ?
r (radius) = 5 cm
Adopting: \pi \approx 3.14

If: We know that the volume of a sphere is V = 4 * \pi *  \frac{r^3}{3}, but we have a hemisphere, so the formula will be half the volume of the hemisphere V =  \frac{1}{2}  * 4 * \pi *  \frac{r^3}{3} &#10;

Formula: (<span>Volume of the hemisphere)
</span>V = \frac{1}{2} * 4 * \pi * \frac{r^3}{3}

Solving:
V = \frac{1}{2} * 4 * \pi * \frac{r^3}{3}
V = \frac{1}{2} * 4 * 3.14 * \frac{5^3}{3}
V = \frac{1}{2} * 4 * 3.14 * \frac{125}{3}
V =  \frac{1570}{6}
\boxed{V_{hemisphere}\approx 261.6\:cm^3}


<span>Now, to find the total volume of the figure, add the values: (cylinder volume + hemisphere volume)
</span>
Volume of the figure = cylinder volume + hemisphere volume
Volume of the figure = 628 cm³ + 261.6 cm³
\boxed{\boxed{Volume\:of\:the\:figure = 1517.6\:cm^3}}\end{array}}\qquad\quad\checkmark
3 0
3 years ago
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We know its tens place and hundred thousandths place but how are they related?
AleksAgata [21]
Well each is still a 3 but with 0s
8 0
3 years ago
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