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3 years ago

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Mathematics

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gayaneshka [121]3 years ago

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In a rhombus ABCD, m∠A = 31°. Point O is a point of intersection of diagonals. Find the measures of the angles of triangle ΔBOC.

MissTica

Check the picture below.

keeping in mind that the diagonals in a rhombus always meet at right-angles, thus all angles at point O are all right-angles.

now, notice that the AC angle bisector, cuts the angle at A in two equal halves, and if you look at the triangle BOC, there is a 15.5° angle, and a right-angle, and well, you already know what the last one is.

4 0

3 years 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.)

7 0

3 years ago

Explain what it means for two variables to be directly related

Karolina [17]

It means that both variables are related in a way that one change in a variable, will result in a change in the other.
There is Directly related and Inversely related. 
Directly is when one change happens to one variable, an equal change will happen to the other. 
Inversely is when one change happens to one variable, an opposite change will happen to the other.

4 0

3 years ago

Read 2 more answers

The y intercept. Plz help I’m so confused and I may ask some questions later as long as I don’t have to pay. Thank you so much I

Masteriza [31]

If you are trying to find the y intercept then you have already found it, it is 4. From that point you would then draw the rest of the slope.

Instead of drawing from (0,0) you would go up 4 units and then do rise over run (7/3) to plot your line.

3 0

3 years ago

Read the proof.

Lostsunrise [7]

A. ∠ACE ≅ ∠ACE

This is correct.