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mr_godi [17]
3 years ago
8

Write six and twenty-three thousandths in standard form

Mathematics
1 answer:
erik [133]3 years ago
7 0
6.023 ..i.think i.am not.gor sure..sorry
You might be interested in
Xy''+2y'-xy by frobenius method
aalyn [17]
First note that x=0 is a regular singular point; in particular x=0 is a pole of order 1 for \dfrac2x.

We seek a solution of the form

y=\displaystyle\sum_{n\ge0}a_nx^{n+r}

where r is to be determined. Differentiating, we have

y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}
y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}

and substituting into the ODE gives

\displaystyle x\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-x\sum_{n\ge0}a_nx^{n+r}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+2\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle \sum_{n\ge0}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r+1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}(n+r)(n+r+1)a_nx^{n+r-1}-\sum_{n\ge2}a_{n-2}x^{n+r-1}=0
\displaystyle r(r+1)a_0x^{r-1}+(r+1)(r+2)a_1x^r+\sum_{n\ge2}\bigg((n+r)(n+r+1)a_n-a_{n-2}\bigg)x^{n+r-1}=0

The indicial polynomial, r(r+1), has roots at r=0 and r=-1. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.

When r=0, we have the recurrence

a_n=\dfrac{a_{n-2}}{(n+1)(n)}

valid for n\ge2. When n=2k, with k\in\{0,1,2,3,\ldots\}, we find

a_0=a_0
a_2=\dfrac{a_0}{3\cdot2}=\dfrac{a_0}{3!}
a_4=\dfrac{a_2}{5\cdot4}=\dfrac{a_0}{5!}
a_6=\dfrac{a_4}{7\cdot6}=\dfrac{a_0}{7!}

and so on, with a general pattern of

a_{n=2k}=\dfrac{a_0}{(2k+1)!}

Similarly, when n=2k+1 for k\in\{0,1,2,3,\ldots\}, we find

a_1=a_1
a_3=\dfrac{a_1}{4\cdot3}=\dfrac{2a_1}{4!}
a_5=\dfrac{a_3}{6\cdot5}=\dfrac{2a_1}{6!}
a_7=\dfrac{a_5}{8\cdot7}=\dfrac{2a_1}{8!}

and so on, with the general pattern

a_{n=2k+1}=\dfrac{2a_1}{(2k+2)!}

So the first indicial root admits the solution

y=\displaystyle a_0\sum_{k\ge0}\frac{x^{2k}}{(2k+1)!}+a_1\sum_{k\ge0}\frac{x^{2k+1}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}
y=\displaystyle \frac{a_0}x\sum_{k\ge0}\frac{x^{2k+1}}{(2k+1)!}+\frac{a_1}x\sum_{k\ge0}\frac{x^{2k+2}}{(2k+2)!}

which you can recognize as the power series for \dfrac{\sinh x}x and \dfrac{\cosh x}x.

To be more precise, the second series actually converges to \dfrac{\cosh x-1}x, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When r=-1, we may seek a second solution of the form

y=cy_1\ln x+x^{-1}\displaystyle\sum_{n\ge0}b_nx^n

where y_1=\dfrac{\sinh x+\cosh x-1}x. Substituting this into the ODE, you'll find that c=0, and so we're left with

y=x^{-1}\displaystyle\sum_{n\ge0}b_nx^n
y=\dfrac{b_0}x+b_1+b_2x+b_3x^2+\cdots

Expanding y_1, you'll see that all the terms x^n with n\ge0 in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form y_2=\dfrac1x. Adding this to y_1, we end up with just \dfrac{\sinh x+\cosh x}x.

This means the general solution for the ODE is

y=C_1\dfrac{\sinh x}x+C_2\dfrac{\cosh x}x
3 0
3 years ago
Find how many gallons it would take to travel 240 miles
svp [43]

Answer:

You will use 8 gallons of gas.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
PLEASEE HELP ME IM VERY CONFUSED ON THIS!!!
Anastasy [175]
SIDE LENGTH OF TRIANGLE: 2.14 inches
SIDE LENGTH OF HEXAGON: 6 inches

To solve this problem, we know that the shapes have equal sides as it states “equilateral triangle”. A triangle has 3 sides and a hexagon has 6 sides. We are told the perimeters are the same so you can set their perimeters equal to each other to solve for x. You would get this : 3(1.4x + 2) = 6(0.5x +2)
With basic algebra you would get x= 5
Then you substitute that value into the length sides of the triangle and hexagon. For the triangle you would approx get 2.14 inches and for the hexagon 6 inches

6 0
2 years ago
Read 2 more answers
Suppose that y varies directly with x and y=10 when x=20. What is y when x=15
Likurg_2 [28]

Answer:

y=kx

subst y=10 and x=20 into the above

<em>1</em><em>0</em><em>=</em><em>k20</em>

<em>k</em><em>=</em><em>1</em><em>0</em><em>/</em><em>2</em><em>0</em>

<em>k</em><em>=</em><em>1</em><em>/</em><em>2</em>

<em>therefore</em><em> </em><em>relationship</em><em>:</em><em> </em><em>y</em><em>=</em><em>1</em><em>/</em><em>2</em><em>x</em>

<em>subst</em><em> </em><em>x</em><em>=</em><em>1</em><em>5</em><em> </em><em>into</em><em> </em><em>the</em><em> </em><em>relationship</em>

<em> </em><em>y</em><em>=</em><em>1</em><em>/</em><em>2</em><em>(</em><em>1</em><em>5</em><em>)</em>

<em>y</em><em>=</em><em>7</em><em>,</em><em>5</em>

Step by step explanation:

  • Step 1: when they say y varies directly with x they mean<em> y is proportional to x</em>
  • step 2: so y=kx where <em>k is the constant</em>
  • step 3: is to substitute <em>y=10</em> and <em>x=20</em> into the above equation y=kx
  • step 4: you will end up with <em>10=k20</em> then divide both sides by 20 so that <em>k becomes the subject of the formula </em>
  • step 5: your answer from the above will be <em>k=10/20 </em>so the relationship is <em>y is directly proportional to 1/2 x </em>what you did here is that you substituted k for 1/2 in the equation in step 3
  • step 6: is to finally substitute x=15 into the equation <em>y=1/2x</em> to finally get your answer <em>y</em><em>=</em><em>7</em><em>,</em><em>5</em><em>.</em>
4 0
3 years ago
Do ∠ABE and ∠DBC share a ray?
Thepotemich [5.8K]

Answer:

yes, it's ray

Step-by-step explanation:

because there are two parallel lines

5 0
3 years ago
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