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3 years ago

What is the equation of the following line (0,0) and (10,-2)

1 answer:

6 0

Okay, the first step to answering this probelm is finding the slope of this line.
The equation for slope is
y₂-y₁
-------
x₂-x₁
Plug the points in
-2 - 0 -2 1
------- = --------- = - ------- (negative 1/5)
10 - 0 10 5
Now we will use the point slope equation
y - y₁ = m(x- x₁)
Lets use the point (10,-2)
Plug in the x,y, and your new slope- then simplify
y +2 = -1/5(x -10)
y + 2 = -1/5x + 2
y = -1/5x

That's your final equation!

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

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

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

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

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

The sum of a number and 4

VLD [36.1K]

Let n be the number.

n + 4

4 0

3 years ago

Read 2 more answers

Are the functions f(x) = (x^2-1)/(x-1) and g(x)= x+1 equal for all x?

Vinvika [58]

$\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\$

$\bf f(x)=\cfrac{x^2-1}{x-1}\implies f(x)=\cfrac{x^2-1^2}{x-1}\implies f(x)=\cfrac{(\underline{x-1})(x+1)}{\underline{x-1}}
\\\\\\
f(x)=x+1\qquad \qquad \qquad \qquad \qquad g(x)=x+1\\\\
-------------------------------\\\\
\textit{they're, kinda, except that, when x = 1}
\\\\\\
g(x)=(1)+1\implies g(x)=2
\\\\\\
f(x)=\cfrac{(1)^2-1}{(1)-1}\implies f(x)=\cfrac{0}{0}\impliedby und efined$

7 0

3 years ago

What is the total finance charge for a $4,250 loan at 13.25% interest compounded monthly for 24 months?

Rom4ik [11]

Since only the principal value, interest rate and interest period are given, we can deduce that "finance charge" only includes the interest to be paid at the end of the term. This can be obtained by subtracting the principal value from the future value which we will solve for.

The future value can be solved by using the following compound interest formula:

Let:
F = Future value
P = Principal value
r = annual interest rate 
n = number of times that interest is compounded per year
t = number of years

F = P(1 + r/n)^nt

Substituting the given values:
F = 4250(1 + 0.1325/12)^(12*2)
F = 5531.54

Subtracting P from F:
Finance charge = 5531.54 - 4250 = 1281.54

Therefore the finance charge is $1,281.54

6 0

3 years ago

Please help, thanks if you do :)

slega [8]

(2,3)
Usbkehisbdhdkd

8 0

2 years ago

Read 2 more answers