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2 years ago

Find the coordinates of the center of the following circle. (x + 3)2 + (y - 6)2 = 24

2 answers:

4 0

<span>the equation of a circle with the center at (h,k0 is given by the equation where the r is obviously the radius of the circle then do (x+3) ^2
(y-6)^2= 24 which shows that the center is = to (-3,6)
</span>

Levart [38]2 years ago

3 0

Answer: (-3,6)

Step-by-step explanation:

We know that the standard equation a circle is given by ":-

$(x-h)^2+(y-k)^2=r^2$ , where r is the center of the circle and (h,k) is the center of the circle.

The given equation of the circle : $(x +3)^2+ (y - 6)^2 = 24$

which is equivalent to $(x -(-3))^2 + (y - 6)^2 = 24$

It implies the coordinates of the center of the given circle =(-3,6)

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What equation can be used to solve for angle C?

matrenka [14]

The formula for calculating the total interior angle in a regular polygon is given as:

total interior angle = (n -2) 180

where n is the number of sides and since there are sides:

total interior angle = 360°

To get this angle, we know that it is the summation of all angles:

360° = angle A + angle C + angle B + angle D ---> 1

Looking closely, we can see that segments BC and AD are parallel which means that:

angle A + angle C = angle B + angle D ---> 2

Substituting equation 2 to 1:

360 = 2 (angle A + angle C)

angle A + angle C = 180

(x + 16) + (6x – 4) = 180

Answer:

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2 years ago

4. In the following diagram, STU, RTP and

Papessa [141]

Yea I don’t know sorry bye

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1 year ago

Consider the discussion in our Devore reading in this unit involving an important distinction between mean and median that uses

Levart [38]

Answer:

Step-by-step explanation:

A trimmed mean is a method of averaging that removes a small designated percentage of the largest and smallest values before calculating the mean. After removing the specified observations, the trimmed mean is found using a standard arithmetic averaging formula. The use of a trimmed mean helps eliminate the influence of data points on the tails that may unfairly affect the traditional mean.

trimmed means provide a better estimation of the location of the bulk of the observations than the mean when sampling from asymmetric distributions;

the standard error of the trimmed mean is less affected by outliers and asymmetry than the mean, so that tests using trimmed means can have more power than tests using the mean.

if we use a trimmed mean in an inferential test , we make inferences about the population trimmed mean, not the population mean. The same is true for the median or any other measure of central tendency.

I can imagine saying the skewness is such-and-such, but that's mostly a side-effect of a few outliers, the fact that the 5% trimmed skewness is such-and-such.

I don't think that trimmed skewness or kurtosis is very much used in practice, partly because

If the skewness and kurtosis are highly dependent on outliers, they are not necessarily useful measures, and trimming arbitrarily solves that problem by ignoring it.

Problems with inconvenient distribution shapes are often best solved by working on a transformed scale.

There can be better ways of measuring or more generally assessing skewness and kurtosis, such as the method above or L-moments. As a skewness measure (mean ? median) / SD is easy to think about yet often neglected; it can be very useful, not least because it is bounded within [?1,1][?1,1].

i expect to see the optimum point in that process at some value between the mean and median.

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1 year ago

Power series of y''+x^2y'-xy=0

Ray Of Light [21]

Assuming we're looking for a power series solution centered around $x=0$ , take

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Substituting into the ODE yields

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0$

The first series starts with a constant term; the second series starts at $x^2$ ; the last starts at $x^1$ . So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a $x^2$ term. We have

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}$

Re-index the first sum to have it start at $n=1$ (to match the the other two sums):

$\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}$

So now the ODE is

$\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0$

Consolidate into one series starting $n=1$ :

$\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0$

Suppose we're given initial conditions $y(0)=a_0$ and $y'(0)=a_1$ (which follow from setting $x=0$ in the power series representations for $y$ and $y'$ , respectively). From the above equation it follows that

$\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}$

Let's first consider what happens when $n=3k-2$ , i.e. $n\in\{1,4,7,10,\ldots\}$ . The recurrence relation tells us that

$a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0$

and so on, so that $a_{3k-2}=0$ except for when $k=1$ .

Now let's consider $n=3k-1$ , or $n\in\{2,5,8,11,\ldots\}$ . We know that $a_2=0$ , and from the recurrence it follows that $a_{3k-1}=0$ for all $k$ .

Finally, take $n=3k$ , or $n\in\{0,3,6,9,\ldots\}$ . We have a solution for $a_3$ in terms of $a_0$ , so the next few terms ( $k=2,3,4$ ) according to the recurrence would be

$a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}$
$a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}$
$a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}$

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for $n=3k$ as

$a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

So the series solution to the ODE is given by

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at $a_0=y(0)=1$ and $a_1=y'(0)=2$ overlaid with the series solution (orange) with $n=3$ and $n=6$ . (Note the rapid convergence.)