Recall that
[tex]\sin(x)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]
Differentiating the power series series for y(x) gives the series for y'(x) :
[tex]y(x)=\displaystyle\sum_{n=0}^\infty a_nx^n \implies y'(x)=\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]
Now, replace everything in the DE with the corresponding power series:
[tex]y'-2xy = 6\sin(3x) \implies[/tex]
[tex]\displaystyle\sum_{n=0}^\infty (n+1)a_{n+1}x^n - 2\sum_{n=0}^\infty a_nx^{n+1} = 6\sum_{n=0}^\infty(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}[/tex]
The series on the right side has no even-degree terms, so if we split up the even- and odd-indexed terms on the left side, the even-indexed [tex](n=2k)[/tex] series should vanish and only the odd-indexed [tex](n=2k+1)[/tex] terms would remain.
Split up both series on the left into even- and odd-indexed series:
[tex]y'(x) = \displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} + \sum_{k=0}^\infty (2k+2)a_{2k+2}x^{2k+1}[/tex]
[tex]-2xy(x) = \displaystyle -2\left(\sum_{k=0}^\infty a_{2k}x^{2k+1} + \sum_{k=0}^\infty a_{2k+1}x^{2k+2}\right)[/tex]
Next, we want to condense the even and odd series:
• Even:
[tex]\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2k+2}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=0}^\infty a_{2k+1}x^{2(k+1)}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2(k-1)+1}x^{2k}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty (2k+1)a_{2k+1}x^{2k} - 2 \sum_{k=1}^\infty a_{2k-1}x^{2k}[/tex]
[tex]=\displaystyle a_1 + \sum_{k=1}^\infty \bigg((2k+1)a_{2k+1} - 2a_{2k-1}\bigg)x^{2k}[/tex]
• Odd:
[tex]\displaystyle \sum_{k=0}^\infty 2(k+1)a_{2(k+1)}x^{2k+1} - 2\sum_{k=0}^\infty a_{2k}x^{2k+1}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2(k+1)}-2a_{2k}\bigg)x^{2k+1}[/tex]
[tex]=\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1}[/tex]
Notice that the right side of the DE is odd, so there is no 0-degree term, i.e. no constant term, so it follows that [tex]a_1=0[/tex].
The even series vanishes, so that
[tex](2k+1)a_{2k+1} - 2a_{2k-1} = 0[/tex]
for all integers k ≥ 1. But since [tex]a_1=0[/tex], we find
[tex]k=1 \implies 3a_3 - 2a_1 = 0 \implies a_3 = 0[/tex]
[tex]k=2 \implies 5a_5 - 2a_3 = 0 \implies a_5 = 0[/tex]
and so on, which means the odd-indexed coefficients all vanish, [tex]a_{2k+1}=0[/tex].
This leaves us with the odd series,
[tex]\displaystyle \sum_{k=0}^\infty \bigg(2(k+1)a_{2k+2}-2a_{2k}\bigg)x^{2k+1} = 6\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}[/tex]
[tex]\implies 2(k+1)a_{2k+2} - 2a_{2k} = \dfrac{6(-1)^k}{(2k+1)!}[/tex]
We have
[tex]k=0 \implies 2a_2 - 2a_0 = 6[/tex]
[tex]k=1 \implies 4a_4-2a_2 = -1[/tex]
[tex]k=2 \implies 6a_6-2a_4 = \dfrac1{20}[/tex]
[tex]k=3 \implies 8a_8-2a_6 = -\dfrac1{840}[/tex]
So long as you're given an initial condition [tex]y(0)\neq0[/tex] (which corresponds to [tex]a_0[/tex]), you will have a non-zero series solution. Let [tex]a=a_0[/tex] with [tex]a_0\neq0[/tex]. Then
[tex]2a_2-2a_0=6 \implies a_2 = a+3[/tex]
[tex]4a_4-2a_2=-1 \implies a_4 = \dfrac{2a+5}4[/tex]
[tex]6a_6-2a_4=\dfrac1{20} \implies a_6 = \dfrac{20a+51}{120}[/tex]
and so the first four terms of series solution to the DE would be
[tex]\boxed{a + (a+3)x^2 + \dfrac{2a+5}4x^4 + \dfrac{20a+51}{120}x^6}[/tex]
ixl area of sectors. I am struggling on this question
Answer:
español :/
Step-by-step explanation:
Answer:
256/5 pi
Step-by-step explanation:
= angle/360 × pi×r^2
= 72/360 × pi × 16^2
= 51.2 pi
= 256/5 pi
The correlation coefficient, r, between the ages of employees, x, and the number of sick days taken per year, y, equals 0.81.
Complete the statement based on the information provided.
The value of r is
✔ positive
and is relatively close to
✔ 1
, so the variables are
✔ closely
associated. It appears that, as the age of an employee increases, the number of sick days taken
✔ increases
.
Answer:
✔ positive
✔ 1
✔ closely
✔ increases
ED2021
Factor the following expressions completely. Show and check all work on your own paper.
x^2+169
If k(x) = 5x - 6, which expression is equivalent to (k+ k)(4)?
Answer:
3h33j333jj3
Step-by-step explanation:b3n3n3nn3n33
Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each quadratic equation with its solution set.
Answer:
2x^2 - 9x -1 = 0
Solution: x = 9 ±√89/4
2x^2 -9x +6 = 0
Solution: 9 ± √33/4
Step-by-step explanation:
Given the equation;2x^2 - 9x -1 = 0
From the quadratic formula;
-b ±√ b^2 -4ac/2a
We have;
x= -(-9) ± √(-9)^2 - -4(2)(-1)/2(2)
x = 9 ±√89/4
Also;
Given the equation: 2x^2 -9x +6 = 0
From the quadratic formula: -b ±√ b^2 -4ac/2a
We have;
x= -(-9) ± √(-9)^2 - 4(2)(6)/2(2)
x= 9 ± √33/4
1.) 2x^2-9x+6 2.) 2x^2-8x+5 3.) 2x^2-9x-1 4.) 2x^2-8x-3
Step-by-step explanation:
let a function F:A➡️B be defined by f(x)=x+1÷2x-1 with A={-1,0,1,2,3,4} and B= {-1,0,4/5,5/7,1,2,3,}.Find the range of f. plzzzz help
Answer:
Range: {-1, 0, 5/7, 4/5, 1, 2}
Step-by-step explanation:
We know that:
f(x) = (x + 1)/(2x - 1)
And:
f: A ⇒ B
where:
A={-1,0,1,2,3,4}
B= {-1,0,4/5,5/7,1,2,3,}
We want to find the range of f(x).
The range of f(x) will be the set of the outputs of f(x) (and because f goes from A to B, we will only take the outputs that belong to B).
Then we only need to evaluate all the values of A in f(x), and see if the output belongs to B.
we have:
f(x) = (x + 1)/(2x - 1)
f(-1) = (-1 + 1)/(2*-1 - 1) = 0 (this does belong to B)
f(0) = (0 + 1)/(2*0 - 1) = -1 (this does belong to B)
f(1) = (1 + 1)/(2*1 - 1) = 2 (this does belong to B)
f(2) = (2 + 1)/(2*2 - 1) = 1 (this does belong to B)
f(3) = (3 + 1)/(2*3 - 1) = 4/5 (this does belong to B)
f(4) = (4 + 1)/(2*4 - 1) = 5/7 (this does belong to B)
So the range of f(x) is the set with all these outputs, which is:
Range: {-1, 0, 5/7, 4/5, 1, 2}
THIS IS NOT A TEST OR ASSESSMENT!! NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! PLEASE EXPLAIN!! Chapter 13
1. What is a conic ? How would you be able to model different conic sections at home(how would you slice a 3D shape to create the conic sections)?
2. How does the equation for the ellipse compare to the equation for a hyperbola? How can you determine the difference?
3. What is the difference between a vertex, a focus, and a directrix?
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Explanation:
1.A cone is a 3-dimensional object created by revolving a line about an axis that intersects that line. This figure is a "double-napped" cone. The point where the revolved line and the axis meet is the a.pex, or vertex, of the cone. Typically, we're concerned with a finite portion of the cone, from the vertex to a base that is a circle in a plane perpendicular to the axis.
A "conic" is a 2-dimensional figure that results from the intersection of a plane and a cone. There are four general categories, named according to the angle the plane makes with the axis and/or the side of the cone. These are illustrated in the attachment.
a circle - the plane of intersection is perpendicular to the axisan ellipse - the plane of intersection is at an angle between 90° and the angle of the side relative to the axis. Both an ellipse and a circle are closed figures.a parabola - the plane of intersection is at the same angle as the side of the cone. A parabola is a one-sided open figure.a hyperbola - The plane of intersection is at an angle between that of the side of the cone and the axis of the cone. The plane will intersect both parts of a double-napped cone producing a double-sided open figure.Producing these at home can be an interesting project. A circle can be made using a compass.
An ellipse can be drawn using a pair of pins and a loop of string. The pins would be placed at the foci of the ellipse, and the string would constrain the drawing instrument (pen or pencil) to have a constant total distance to the two foci.
A parabola can be drawn on graph paper using coordinates derived from an equation for it. It can also be drawn using a compass and a set square by plotting points that are equidistant from the focus and a line that is called the directrix. If you have a physical cone-shaped object, you can cut it at an angle that will produce a parabola.
A hyperbola can be drawn on graph paper from an equation. It can also be drawn using a compass by plotting points that have a constant difference in their distance to the two foci, or by plotting points whose ratio of distance to focus and directrix is a constant. A physical cone-shaped object can be cut to produce a hyperbola.
__
2.The general form equation for a conic is ...
Ax² +Bxy +Cy² +Dx +Ey +F = 0
Usually, we're concerned with conics that have axes parallel to the coordinate axes, so B=0. The equation of an ellipse has A and C with the same sign. The equation of a hyperbola has A and C with opposite signs,
In standard form, the equations for figures centered at the origin are ...
ellipse: x²/a +y²/b = 1hyperbola: x²/a -y²/b = 1 (opens horizontally)hyperbola: y²/a -x²/b = 1 (opens vertically)__
3.The vertex of a conic is an extreme point on the (major) axis of the conic. The focus is a point used in the definition of the conic. The focus is "inside" the curve, on the axis of symmetry. The directrix is a line used in the definition of the conic. The directrix is "outside" the curve, perpendicular to the axis. The second attachment shows these for a parabola.
what is the union of these two sets? E={-1,0,4,5,6,7} G={-2,-1,1,2,3,8}
Answer:
U={-2,-1,0,1,2,3,4,5,6,7,8}
Help me plz help me plz
Answer: 4 13/30 cups
Step-by-step explanation:
Since Lila used 1 2/5 times as much lemonade as Naomi did (3 1/6 cups), we have to multiply 1 2/5 by 3 1/6:
1 2/5 ⋅ 3 1/6 = ?
19/6 ⋅ 7/5 = 133/30
133/30 = 4 13/30
4 13/30 cups
A concave polygon can never be classified as a regular polygon true or false??? Need answer ASAP please
Answer:
Regular Polygons are never concave by definition.
Step-by-step explanation:
proving lines parallel!!! please help
Answer:
B.
Step-by-step explanation:
Since C and A are parallel
and B is perpendicular to C
then B is also perpendicular to A
meaning they have the same angle so 90 Degrees.
The number of dollars in x quarters
Answer:
There are four quarters in one dollar, so 4x quarters in x dollarsFind the length of X
Answer:
[tex] x = 8\sqrt{2} [/tex]
Step-by-step explanation:
Leg = x
Hypotenuse = 16
[tex] x\sqrt{2} = 16 [/tex]
[tex] x = \dfrac{16}{\sqrt{2}} [/tex]
[tex] x = \dfrac{16}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} [/tex]
[tex] x = \dfrac{16\sqrt{2}}{2} [/tex]
[tex] x = 8\sqrt{2} [/tex]
During a 1966 Tabiona High School track meet, Levere ran the 100 yard dash in
10.63 seconds. Ross took second with a time of 10.98 seconds.
a. Levere’s time was _______% shorter than Ross’.
b. Ross’ time was _______% longer than Levere’s.
c. Levere’s time was _______% of Ross’.
Answer:
a) 3.19
b) 3.29
c) 96.81
Step-by-step explanation:
Question a:
Levere's: 10.63s
Ross: 10.98s
10.98 - 10.63 = 0.35s shorter than 10.98s, so:
0.35*100%/10.98 = 3.19% shorter.
Question b:
35s longer than 10.63s, so:
0.35*100%/10.63 = 3.29% longer.
Question c:
3.19% shorter, so 100 - 3.19 = 96.81% of Ross.
A dinner mint costs 85¢ and a toffee costs 73¢. What is the cost of both sweets rounded to the nearest dollar?
Answer:
85 rounded to the nearest dollar would be $1. 73 rounded to the nearest dollar would also be $1
Step-by-step explanation:
What is the solution to the system of equations? 5x-2y=-16 4x-5y=-23
Answer:
The solution set is {-2, 3}.
Step-by-step explanation:
We can do this by elimination after manipulating the 2 equations so that the coefficients of y will disappear after addition:
5x - 2y = -16 Multiply this by -5:
-25x + 10y = 80 ...........(A)
4x - 5y = -23 Multiply this by 2:
8x - 10y = -46..............(B)
Now add A and B:
-17x = 34
x = 34/-17
x = -2.
Now substitute x = -2 into the first original equation:
5(-2) - 2y = -16
-2y = -16 +10 = -6
y = -6/-2
y = 3.
Confirm these results by substitution in the second original equation:
4(-2) - 5(3)
= - 8 - 15 = -23.
Checks OK.
AM and CM
BM and BM
AB and CB
These are variables on your graph
What is an explicit formula for the geometric sequence -64,16,-4,1,... where the first term should be f(1).
Answer:
[tex]a_{n} = -64(-\frac{1}{4})^{n-1}[/tex]
it seems like the first term is -64, so lets write the formula accordingly:
a_n = a1(r)^(n-1)
where 'n' is the number of terms
a1 is the first term of the sequence
'r' is the ratio
the ratio is [tex]-\frac{1}{4}[/tex] because -64 * [tex]-\frac{1}{4}[/tex] = 16 and so on...
the explicit formula is :
[tex]a_{n}[/tex] = [tex]-64(-\frac{1}{4} )^{n-1}[/tex]
You play a game where you roll a single die. You pay $1 to play, and the payouts are $0.50 if you roll an
even number, $2 if you roll a 1, and $1 if you roll a 3 or 5.
2. What are the odds for winning money if you play this game? Show your work and Explain.
dan
3. What is the expected value of this game? Show your work and Explain what the results mean.
All you do is first you ha
twice the difference of a number and 8 is 6. use the variable x for the unknown number.
Answer:
11
Step-by-step explanation:
Unknown number = x
If twice the difference of x and 8 is 6:
2(x-8) = 6
2x-16 = 6
2x = 6 + 16
2x = 22
x = 22/2
x = 11
Answer from Gauthmath
g Two different factories named A and B both produce an automobile part. If a part came from A, the probability that the part is defective is .04. If the part came from B, the probability that it is defective is .05. In a sample of 180 parts, 100 came from A and 80 came from B. (a) What is the probability that a part chosen at random (from the sample) was defective
Answer:
0.0444 = 4.44% probability that a part chosen at random (from the sample) was defective.
Step-by-step explanation:
Probability of a defective part:
0.04 of [tex]\frac{100}{180}[/tex], that is, coming from A.
0.05 of [tex]\frac{80}{180}[/tex], that is, coming from B. So
[tex]p = 0.04\frac{100}{180} + 0.05\frac{80}{180} = \frac{0.04*100 + 0.05*80}{180} = 0.0444[/tex]
0.0444 = 4.44% probability that a part chosen at random (from the sample) was defective.
AABC is reflected across the x-axis and then translated 4 units up to create AA'BC. What are the coordinates of the vertices of AABC?
What error, if any, did Noah make?
Answer:
breathing, jk buddy
Step-by-step explanation:
Suppose a deck of cards contains 13 cards:
5 green cards numbered 1-5, 4 red cards numbered 1-4, and 4 blue cards numbered 1-4.
For 3.1-3.3, 5 draws are made without replacement. X is the number of green cards drawn and Y is the number of red cards drawn. Z is the sum of the numbers on the tickets.
G1 = first card is green
G2 = second card is green
Enter the probability as a fraction.
P(at least one green) = ______.
Answer:
[tex]P(G_1) = \frac{5}{13}[/tex]
[tex]P(G_2) = \frac{1}{3}[/tex]
[tex]P(X \ge 1) = \frac{25}{39}[/tex]
Step-by-step explanation:
Given
[tex]G = 5[/tex]
[tex]R = 4[/tex]
[tex]B = 4[/tex]
[tex]n = 13[/tex]
Solving (a): [tex]P(G_1)[/tex]
This is calculated as:
[tex]P(G_1) = \frac{G}{n}[/tex]
[tex]P(G_1) = \frac{5}{13}[/tex]
Solving (b): [tex]P(G_2)[/tex]
This is calculated as:
[tex]P(G_2) = \frac{G - 1}{n - 1}[/tex] -- this is so because the selection is without replacement
[tex]P(G_2) = \frac{5 - 1}{13 - 1}[/tex]
[tex]P(G_2) = \frac{4}{12}[/tex]
[tex]P(G_2) = \frac{1}{3}[/tex]
Solving (c): [tex]P(X \ge 1)[/tex]
Using the complement rule, we have:
[tex]P(X \ge 1) = 1 - P(X = 0)[/tex]
To calculate [tex]P(X = 0)[/tex], we have:
[tex]G = 5[/tex] --- Green
[tex]G' = 8[/tex] ---- Not green
The probability that both selections are not green is:
[tex]P(X = 0) = P(G'_1) * P(G'_2)[/tex]
So, we have:
[tex]P(X = 0) = \frac{G'}{n} * \frac{G'-1}{n-1}[/tex]
[tex]P(X = 0) = \frac{8}{13} * \frac{8-1}{13-1}[/tex]
[tex]P(X = 0) = \frac{8}{13} * \frac{7}{12}[/tex]
Simplify
[tex]P(X = 0) = \frac{2}{13} * \frac{7}{3}[/tex]
[tex]P(X = 0) = \frac{14}{39}[/tex]
Recall that:
[tex]P(X \ge 1) = 1 - P(X = 0)[/tex]
[tex]P(X \ge 1) = 1 - \frac{14}{39}[/tex]
Take LCM
[tex]P(X \ge 1) = \frac{39 -14}{39}[/tex]
[tex]P(X \ge 1) = \frac{25}{39}[/tex]
Help me plz help me plz plz
Im sorry I don't know the answer to the question
Hi! I'd appreciate it if you could help me on this question. The question I need help with is question 42. Thank you If you could help me!!
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Answer:
19 hours
Step-by-step explanation:
Add up the numbers:
3×3.5 +2×2.0 +4.5 = 19
Jacob trains 19 hours per week.
How much bigger is the Sum of first 50 even numbers than the sum of first 50 odd numbers?
Answer:
50
Step-by-step explanation:
Sum Even numbers
n = 50
d = 2
a1 = 2
The last number is
an = a1 + (n-1)d
an = 2 + (50 - 1)*2
an = 2 + 49 * 2
an = 2 + 98
an = 100
Sum of the even numbers
Sum = (a1 + a50)*n/ 2
Sum = (2 + 100)*50/2
sum = 102 * 25
sum = 2550
Sum of the first 50 odd numbers
a1 = 1
n = 50
d = 2
l = ?
Find l
l = a1 + (n - 1)*2
l = 1 + 49*2
l = 99
Sum
Sum = (1 + 99)*50/2
Sum = 2500
The difference and answer is 2550 - 2500 = 50
Verify that the indicated function y = ϕ(x) is an explicit solution of the given first-order differential equation. (y − x)y' = y − x + 18; y = x + 6 √(x + 4)
When y = x + 6√(x + 4) y'=_________
Thus in terms of x, (y-x)y'=________
y-x+18=________
If y = x + 6√(x + 4), then
y' = 1 + 3/√(x + 4)
Substituting y and y' into the DE gives
(y - x) y' = (x + 6√(x + 4) - x) (1 + 3/√(x + 4))
… = 6√(x + 4) (1 + 3/√(x + 4))
… = 6√(x + 4) + 18
on the left side, while on the right you get
y - x + 18 = x + 6√(x + 4) - x + 18
… = 6√(x + 4) + 18
so both sides match and the given function is indeed a solution to the DE.
A one lane highway runs through a tunnel in the shape of one half a sine curve cycle
The sine curve equation, y = 10·sin(x·π/24), that models the entrance of the
tunnel with a cross section that is the shape of half of a sine curve and the
height of the tunnel at the edge of the road, (approximately 7.07 ft.) are
found by applying the following steps
(a) The equation for the sine curve is y = 10·sin(x·π/24)
(b) The height of the tunnel at the edge of the road is approximately 7.07 feet
The reason for the above answers are presented as follows;
(a) From a similar question posted online, the missing part of the question
is, what is the height of the tunnel at the edge of the road
The known parameters;
The shape of the tunnel = One-half sine curve cycle
The height of the road at its highest point = 10 ft.
The opening of the tunnel at road level = 24 ft.
The unknown parameter;
The equation of the sine curve that fits the opening
Method;
Model the sine curve equation of the tunnel using the general equation of a sine curve;
The general equation of a sine curve is y = A·sin(B·(x - C) + D
Where;
y = The height at point x
A = The amplitude = The distance from the centerline of the sine wave to the top of a crest
Therefore;
The amplitude, A = The height of half the sine wave = The height of the tunnel = 10 ft.
D = 0, C = 0 (The origin, (0, 0) is on the left end, which is the central line)
The period is the distance between successive points where the curve passes through the center line while rising to a crest
Therefore
The period, T = 2·π/B = 2 × Opening at the road level = 2 × 24 ft. = 48 ft.
T = 48 ft.
We get;
48 = 2·π/B
B = 2·π/48 = π/24
By plugging in the values for A, B, C, and D, we get;
y = 10·sin((π/24)·(x - 0) + 0 = 10·sin(x·π/24)
The equation of the sine curve that fits the opening is y = 10·sin(x·π/24)
(b) The height of the tunnel at the edge of the road is given by substituting
the value of x at the edge of the road into the equation for the sine curve
as follows;
The width of the shoulders = 6 feet
∴ At the edge of the road, x = 0 + 6ft = 6 ft., and 6 ft. + 12 ft. = 18 ft.
Therefore, we get;
y = 10 × sin(6·π/24) = 10 × sin(π/4) = 5×√2
y = 10 × sin(18·π/24) = 10 × sin(3·π/4) = 5×√2
The height of the, y, tunnel at the edge of the road where, x = 6, and 18 is y = 5·√2 feet ≈ 7.07 ft.
Learn more about the sine curve here;
https://brainly.com/question/3827606
Muka saved 476.60. He gave Kelvin 429.10 and bought T-Shirt for 432.05, how much money he has left over
