By letting
![y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}](https://tex.z-dn.net/?f=y%20%3D%20%5Cdisplaystyle%20%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20c_n%20x%5E%7Bn%2Br%7D)
we get derivatives
![y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}](https://tex.z-dn.net/?f=y%27%20%3D%20%5Cdisplaystyle%20%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20%28n%2Br%29%20c_n%20x%5E%7Bn%2Br-1%7D)
![y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}](https://tex.z-dn.net/?f=y%27%27%20%3D%20%5Cdisplaystyle%20%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20%28n%2Br%29%20%28n%2Br-1%29%20c_n%20x%5E%7Bn%2Br-2%7D)
a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to
![5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0](https://tex.z-dn.net/?f=5r%28r-1%29%20c_0%20x%5E%7Br-1%7D%20%2B%20%5Cdisplaystyle%20%5Csum_%7Bn%3D1%7D%5E%5Cinfty%20%5Cbigg%28%20%28n%2Br%2B1%29%20c_n%20%2B%20%28n%20%2B%20r%20%2B%201%29%20%285n%20%2B%205r%20%2B%201%29%20c_%7Bn%2B1%7D%20%5Cbigg%29%20x%5E%7Bn%2Br%7D%20%3D%200)
Examine the lowest degree term
, which gives rise to the indicial equation,
![5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0](https://tex.z-dn.net/?f=5r%20%28r%20-%201%29%20%2B%20r%20%3D%200%20%5Cimplies%205r%5E2%20-%204r%20%3D%20r%20%285r%20-%204%29%20%3D%200)
with roots at r = 0 and r = 4/5.
b) The recurrence for the coefficients
is
![(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}](https://tex.z-dn.net/?f=%28k%2Br%2B1%29%20c_k%20%2B%20%28k%20%2B%20r%20%2B%201%29%20%285k%20%2B%205r%20%2B%201%29%20c_%7Bk%2B1%7D%20%3D%200%20%5Cimplies%20c_%7Bk%2B1%7D%20%3D%20-%5Cdfrac%7Bc_k%7D%7B5k%2B5r%2B1%7D)
so that with r = 4/5, the coefficients are governed by
![c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}](https://tex.z-dn.net/?f=c_%7Bk%2B1%7D%20%3D%20-%5Cdfrac%7Bc_k%7D%7B5k%2B5%7D%20%5Cimplies%20%5Cboxed%7Bg%28k%29%20%3D%20-%5Cdfrac1%7B5k%2B5%7D%7D)
c) Starting with
, we find
![c_1 = -\dfrac{c_0}5 = -\dfrac15](https://tex.z-dn.net/?f=c_1%20%3D%20-%5Cdfrac%7Bc_0%7D5%20%3D%20-%5Cdfrac15)
![c_2 = -\dfrac{c_1}{10} = \dfrac1{50}](https://tex.z-dn.net/?f=c_2%20%3D%20-%5Cdfrac%7Bc_1%7D%7B10%7D%20%3D%20%5Cdfrac1%7B50%7D)
so that the first three terms of the solution are
![\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum_%7Bn%3D0%7D%5E2%20c_n%20x%5E%7Bn%20%2B%204%2F5%7D%20%3D%20%5Cboxed%7Bx%5E%7B4%2F5%7D%20-%20%5Cdfrac15%20x%5E%7B9%2F5%7D%20%2B%20%5Cfrac1%7B50%7D%20x%5E%7B13%2F5%7D%7D)
The student is correct. Due to the meaning of perimeter which is all sides added up to make a number total, it is possible for a square and a rectangle to have the same perimeter if the numbers add up.
Answer:
<em>19800 seconds, or 330 minutes, or 5 hours + 30 minutes</em>
Step-by-step explanation:
<u>Number Permutations</u>
We know the phone number has 7 digits, 4 of which are known by Mark. This leaves him 3 digits to guess with. We also know the last one is not zero. The number can be represented as
XXY
Where X can be any digit from 0 to 9 and Y can be any digit from 1 to 9. The first two can be combined in 10x10 ways, and the last one can be of 9 ways, this gives us 10x10x9 = 900 possible permutations.
If each possible number takes him 22 seconds, every possibility will need
22x900=19800 seconds, or 330 minutes, or 5 hours + 30 minutes
<h2>
Explanation:</h2>
In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.
So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.
So let's name the vertices as:
![A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)](https://tex.z-dn.net/?f=A%28-1%2C3%29%20%5C%5C%20%5C%5C%20B%281%2C5%29%20%5C%5C%20%5C%5C%20C%285%2C1%29%20%5C%5C%20%5C%5C%20D%283%2C-1%29)
First pair of opposite sides:
<u>Slope:</u>
![\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}](https://tex.z-dn.net/?f=%5Ctext%7BFor%20AB%7D%3A%20%5C%5C%20%5C%5C%20m%3D%5Cfrac%7B5-3%7D%7B1-%28-1%29%7D%3D1%20%5C%5C%20%5C%5C%20%5C%5C%20%5Ctext%7BFor%20CD%7D%3A%20%5C%5C%20%5C%5C%20m%3D%5Cfrac%7B1-%28-1%29%7D%7B5-3%7D%3D1%20%5C%5C%20%5C%5C%20%5C%5C%20%5Ctext%7BSo%20AB%20and%20CD%20are%20parallel%7D)
Second pair of opposite sides:
<u>Slope:</u>
![\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}](https://tex.z-dn.net/?f=%5Ctext%7BFor%20BC%7D%3A%20%5C%5C%20%5C%5C%20m%3D%5Cfrac%7B1-5%7D%7B5-1%7D%3D-1%20%5C%5C%20%5C%5C%20%5C%5C%20%5Ctext%7BFor%20AD%7D%3A%20%5C%5C%20%5C%5C%20m%3D%5Cfrac%7B-1-3%7D%7B3-%28-1%29%7D%3D-1%20%5C%5C%20%5C%5C%20%5C%5C%20%5Ctext%7BSo%20BC%20and%20AD%20are%20parallel%7D)
So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:
![d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%28y_%7B2%7D-y_%7B1%7D%29%5E2%2B%28x_%7B2%7D-x_%7B1%7D%29%5E2%7D%20%5C%5C%20%5C%5C%20%5C%5C%20Diagonal%20%5C%20BD%3A%20%5C%5C%20%5C%5C%20d%3D%5Csqrt%7B%285-%28-1%29%29%5E2%2B%281-3%29%5E2%7D%3D2%5Csqrt%7B10%7D%20%5C%5C%20%5C%5C%20%5C%5C%20Diagonal%20%5C%20AC%3A%20%5C%5C%20%5C%5C%20d%3D%5Csqrt%7B%283-1%29%5E2%2B%28-5-1%29%5E2%7D%3D2%5Csqrt%7B10%7D%20%5C%5C%20%5C%5C%20%5C%5C)
So the diagonals measure the same, therefore this is a rectangle.
Answer:
<h2>58°</h2>
Step-by-step explanation:
We will use the tangent function since we know the opposite and adjacent sides.
Tangent = opposite/adjacent
Tan(e) = 16/10
Tan(e) = 1.6
Use the inverse tangent function to find the angle.
Arctan (1.6) = 57.9946168
Rounding this we get: 58°