The sum we want is
[tex]\displaystyle \sum_{n=0}^\infty \frac{(-1)^{T_n}}{(2n+1)^2} = 1 - \frac1{3^2} - \frac1{5^2} + \frac1{7^2} + \cdots[/tex]
where [tex]T_n=\frac{n(n+1)}2[/tex] is the n-th triangular number, with a repeating sign pattern (+, -, -, +). We can rewrite this sum as
[tex]\displaystyle \sum_{k=0}^\infty \left(\frac1{(8k+1)^2} - \frac1{(8k+3)^2} - \frac1{(8k+7)^2} + \frac1{(8k+7)^2}\right)[/tex]
For convenience, I'll use the abbreviations
[tex]S_m = \displaystyle \sum_{k=0}^\infty \frac1{(8k+m)^2}[/tex]
[tex]{S_m}' = \displaystyle \sum_{k=0}^\infty \frac{(-1)^k}{(8k+m)^2}[/tex]
for m ∈ {1, 2, 3, …, 7}, as well as the well-known series
[tex]\displaystyle \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = -\frac{\pi^2}{12}[/tex]
We want to find [tex]S_1-S_3-S_5+S_7[/tex].
Consider the periodic function [tex]f(x) = \left(x-\frac12\right)^2[/tex] on the interval [0, 1], which has the Fourier expansion
[tex]f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi nx)}{n^2}[/tex]
That is, since f(x) is even,
[tex]f(x) = a_0 + \displaystyle \sum_{n=1}^\infty a_n \cos(2\pi nx)[/tex]
where
[tex]a_0 = \displaystyle \int_0^1 f(x) \, dx = \frac1{12}[/tex]
[tex]a_n = \displaystyle 2 \int_0^1 f(x) \cos(2\pi nx) \, dx = \frac1{n^2\pi^2}[/tex]
(See attached for a plot of f(x) along with its Fourier expansion up to order n = 10.)
Expand the Fourier series to get sums resembling the [tex]S'[/tex]-s :
[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{k=0}^\infty \frac{\cos(2\pi(8k+1) x)}{(8k+1)^2} + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+2) x)}{(8k+2)^2} + \cdots \right. \\ \,\,\,\, \left. + \sum_{k=0}^\infty \frac{\cos(2\pi(8k+7) x)}{(8k+7)^2} + \sum_{k=1}^\infty \frac{\cos(2\pi(8k) x)}{(8k)^2}\right)[/tex]
which reduces to the identity
[tex]\pi^2\left(\left(x-\dfrac12\right)^2-\dfrac{21}{256}\right) = \\\\ \cos(2\pi x) {S_1}' + \cos(4\pi x) {S_2}' + \cos(6\pi x) {S_3}' + \cos(8\pi x) {S_4}' \\\\ \,\,\,\, + \cos(10\pi x) {S_5}' + \cos(12\pi x) {S_6}' + \cos(14\pi x) {S_7}'[/tex]
Evaluating both sides at x for x ∈ {1/8, 3/8, 5/8, 7/8} and solving the system of equations yields the dependent solution
[tex]\begin{cases}{S_4}' = \dfrac{\pi^2}{256} \\\\ {S_1}' - {S_3}' - {S_5}' + {S_7}' = \dfrac{\pi^2}{8\sqrt 2}\end{cases}[/tex]
It turns out that
[tex]{S_1}' - {S_3}' - {S_5}' + {S_7}' = S_1 - S_3 - S_5 + S_7[/tex]
so we're done, and the sum's value is [tex]\boxed{\dfrac{\pi^2}{8\sqrt2}}[/tex].
A 2-pack of clay pots costs $9.30. What is the unit price?
Answer:
$4.75
Step-by-step explanation:
To find the unit price (price for one), just divide $9.30 by 2. You get that one clay pot (the price for one unit) is $4.75.
Hopefully this helps- let me know if you have any questions!
Answer my question #2
Answer:
Ans is mean .because mean means average.
what is 5(m-2) equil to
Answer:
5m - 10
Step-by-step explanation:
Use distributive property;
5(m - 2)
(5 x m) (5 x -2)
5m - 10
There is a bag filled with 3 blue and 5 red marbles.
A marble is taken at random from the bag, the colour is noted and then it is replaced.
Another marble is taken at random.
What is the probability of getting at least 1 red?
Answer:
1/4 is the correct answer
Step-by-step explanation:
Even after replacing the marble the propability will not chance.
Determine all the zeros for the function f(x) = (x2 + 3x - 10)(x - 4).
Answer:
(x−4)(x−2)(x+5)
Step-by-step explanation:
the length of a rectagle is 6 ft longer than its width. if the perimeter of the rectangle is 64 ft, find its length and width
Answer:
19 and 13
Step-by-step explanation:
width = x
length = x + 6
2x + 2(x + 6) = 64
2x + 2x + 12 = 64
4x = 64 - 12
4x = 52
x = 13
x + 6 = 19
Answer:
19 and 13
Step-by-step explanation:
has parallel sides but is not a trapezoid
Answer:
Parallelogram
Step-by-step explanation:
Instead of one pair of parallel sides like in a trapezoid, a parallelogram has two pairs of opposite sides (for example a rectangle)
-5x+y=-10 and -4x-y=-26
[tex]\left\{\begin{matrix}x=4\\y=10\\\end{matrix}\right.[/tex]
Step-by-step explanation:Solve the equation[tex]\left\{\begin{matrix}-5x+y=-10\\-4x-y=-26\\\end{matrix}\right.[/tex]
_____________________________________
Add the two equations[tex]-5x+y+(-4x-y)=-10+(-26)[/tex]
____________________________
Remove parentheses[tex]-5x+y-4x-y=-10-26[/tex]
_________________________________________________
Cancel one variable[tex]-5x-4x=-10-26[/tex]
___________________________________________
Combine like terms[tex]-9x=-10-26[/tex]
______________________________________________
Calculate the sum or difference[tex]-9x=-36[/tex]
________________________________________________
Divide both sides of the equation by the coefficient of variable
[tex]x=\frac{-36}{-9}[/tex]
____________________________________________________
Determine the sign for multiplication or division
[tex]x=\frac{36}{9}[/tex]
_________________________________________
Cross out the common factor
[tex]x=4[/tex]
_____________________________________________Step-by-step explanation:Substitute into one of the equations[tex]-4\times4-y=-26[/tex]
____________________________________________________
Calculate the product or quotient
[tex]-16-y=-26[/tex]
______________________________________________________
Rearrange variables to the left side of the equation
[tex]-y=-26+16[/tex]
_____________________________________________
Calculate the sum or difference
[tex]-y=-10[/tex]
_________________________________________-
Divide both sides of the equation by the coefficient of variable
[tex]y=10[/tex]
I hope this helps you
:)
Write the equation of the trigonometric graph.
Answer(s):
[tex]\displaystyle y = 4sin\:(2x + \frac{\pi}{2}) \\ y = 4cos\:2x[/tex]
Step-by-step explanation:
[tex]\displaystyle y = Asin(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \hookrightarrow \boxed{-\frac{\pi}{4}} \hookrightarrow \frac{-\frac{\pi}{2}}{2} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 4[/tex]
OR
[tex]\displaystyle y = Acos(Bx - C) + D \\ \\ Vertical\:Shift \hookrightarrow D \\ Horisontal\:[Phase]\:Shift \hookrightarrow \frac{C}{B} \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \\ Amplitude \hookrightarrow |A| \\ \\ Vertical\:Shift \hookrightarrow 0 \\ Horisontal\:[Phase]\:Shift \hookrightarrow 0 \\ Wavelength\:[Period] \hookrightarrow \frac{2}{B}\pi \hookrightarrow \boxed{\pi} \hookrightarrow \frac{2}{2}\pi \\ Amplitude \hookrightarrow 4[/tex]
You will need the above information to help you interpret the graph. First off, keep in mind that although this looks EXACTLY like the cosine graph, if you plan on writing your equation as a function of sine, then there WILL be a horisontal shift, meaning that a C-term will be involved. As you can see, the photograph on the right displays the trigonometric graph of [tex]\displaystyle y = 4sin\:2x,[/tex]in which you need to replase "cosine" with "sine", then figure out the appropriate C-term that will make the graph horisontally shift and map onto the cosine graph [photograph on the left], accourding to the horisontal shift formula above. Also keep in mind that the −C gives you the OPPOCITE TERMS OF WHAT THEY REALLY ARE, so you must be careful with your calculations. So, between the two photographs, we can tell that the sine graph [photograph on the right] is shifted [tex]\displaystyle \frac{\pi}{4}\:unit[/tex]to the right, which means that in order to match the cosine graph [photograph on the left], we need to shift the graph BACKWARD [tex]\displaystyle \frac{\pi}{4}\:unit,[/tex]which means the C-term will be negative, and by perfourming your calculations, you will arrive at [tex]\displaystyle \boxed{-\frac{\pi}{4}} = \frac{-\frac{\pi}{2}}{2}.[/tex]So, the sine graph of the cosine graph, accourding to the horisontal shift, is [tex]\displaystyle y = 4sin\:(2x + \frac{\pi}{2}).[/tex]Now, with all that being said, in this case, sinse you ONLY have a graph to wourk with, you MUST figure the period out by using wavelengths. So, looking at where the graph WILL hit [tex]\displaystyle [-1\frac{1}{4}\pi, 0],[/tex]from there to [tex]\displaystyle [-\frac{\pi}{4}, 0],[/tex]they are obviously [tex]\displaystyle \pi\:units[/tex]apart, telling you that the period of the graph is [tex]\displaystyle \pi.[/tex]Now, the amplitude is obvious to figure out because it is the A-term, but of cource, if you want to be certain it is the amplitude, look at the graph to see how low and high each crest extends beyond the midline. The midline is the centre of your graph, also known as the vertical shift, which in this case the centre is at [tex]\displaystyle y = 0,[/tex]in which each crest is extended four units beyond the midline, hence, your amplitude. So, no matter how far the graph shifts vertically, the midline will ALWAYS follow.
I am delighted to assist you at any time.
This is a img thank you
Answer: A = and then below is incorrect then state how to find the correct answer and why it is correct.
Step-by-step explanation: hope it helps
jill has a small treasure box
Trigonometric equations
4sin^2(theta) + 4 = 5
Answer:
Θ = (π/6) + πn
Θ = (5π/6) + πn
Step-by-step explanation:
4sin²Θ + 4 = 5
-4 -4
4sin²Θ = 1
÷4 ÷4
sin²Θ = (1/4)
√sin²Θ = √(1/4)
sinΘ = (1/2), (-1/2)
-------------------------
Θ = arcsin (1/2)
Θ = (π/6)
to find the quadrant subtract π
Θ = π - (π/6)
Θ = (5π/6)
Find the period
2π / |b|
b = 1
2π/1 = 2π
The sin Θ function is 2π, so values will repeat 2π in both directions.
Θ = (π/6) + 2πn (n is the variable)
Θ = (5π/6) + 2πn
-------------------------------------------------------------------------------------------------------
sin Θ = (-1/2)
Θ = arcsin (-1/2)
Θ = (-π/6)
To find the second function add π
Θ = 2π + (π/6) + π
Θ = (7π/6)
Find the period
2π/|b|
2π/1
2π
(-π/6) + 2π
2π 6 π
----- × ----- - -----
1 6 6
Θ = (11π/6) will repeat every 2π in both directions
-------------------------------------------------------------------------------------------------------
Θ = (π/6) + 2πn
Θ = (5π/6) + 2πn
Θ = (7π/6) + 2πn
Θ = (11π/6) + 2πn
(π/6) + π = (7π/6)
(5π/6) + π = (11π/6)
Θ = (π/6) + πn
Θ = (5π/6) + πn
----------------------------------------------------------------------------------------------------------
I hope this helps!
3
Select all the correct answers.
Which three statements are true as they relate to supply and demand?
As supply rises, prices generally decrease.
o As demand decreases, costs generally increase.
As supply decreases, prices increase.
The average rate of change describes how much a quantity changes as price increases.
o As demand rises, the price of the product decreases.
Answer:
A, C, and D
Step-by-step explanation:
Let's analyze each option:
A) As supply rises, prices generally decrease. This is true as supply and demand have an inverse relationship. The supply curve has a positive slope, showing that as the quantity supplied increases, the price increases. So, shifting the curve to the right does decrease the equilibrium price (assuming the demand curve doesn't fluctuate).
B) As demand decreases, costs generally increase. This is false because a demand curve shows that as the quantity of demand increases, the price decreases for the product. So, shifting the curve to the left only decreases the equilibrium price (assuming the supply curve doesn't fluctuate).
C) As supply decreases, prices increase. This is true because the supply curve has a positive slope, showing that as the quantity supplied increases, the price increases. So, shifting the curve to the left does increase the equilibrium price (assuming the demand curve doesn't fluctuate).
D) The average rate of change describes how much a quantity changes as price increases. This is true because the supply curve measures how much the quantity supplied increases as the price increases, while the demand curve measures how much the quantity demanded increases as the price decreases.
E) As demand rises, the price of the product decreases. This is false because a demand curve shows that as the quantity of demand increases, the price decreases for the product. So, shifting the curve to the right only increases the equilibrium price (assuming the supply curve doesn't fluctuate).
I've attached an example graph to help you visualize all of these factors.
Select the most appropriate unit for the situation
Rate of filling a buck with water
Feet / minute
Square feet /minute
Cubic feet / minute
Answer: Is Square feet/minute if there is two of them the second one is Cubic feet/minute
Step-by-step explanation: feet/minute would cover anything linear {forward, backward, left, right}
square feet/minute would cover two-dimensional areas {walls, floors, sides of buildings, etc.}
cubic feet/minute would cover three dimensional situations {swimming pools, buckets, containers, etc.}
Is the table a linear or exponential function?
Answer:
exponential
Step-by-step explanation:
there is a multiplier of 8 between each number
Look at the image below. What is the area of the parallelogram? by Middle School
Apply Pythagorean theorem
[tex]\\ \rm\rightarrowtail B^2=2.2^2-2^2[/tex]
[tex]\\ \rm\rightarrowtail B^2=5-4[/tex]
[tex]\\ \rm\rightarrowtail B^2=1[/tex]
[tex]\\ \rm\rightarrowtail B=1[/tex]
Base=1+3=4Area:-
Base×Height4(2)8units^2Answer:6
Step-by-step explanation:
I did the test
Find P (Freshman(Girl) Hint: P(A and B)
P(B)
Classroom of Students
Boys Girls
Freshman 4 4 8
Sophomore 5 7 12
Junior 2 3 5
Senior 1
4 5
12 18
Reduce your fraction to lowest terms.
Enter the number that belongs in the green box.
Answer:
[tex]\sf \dfrac29[/tex]
Step-by-step explanation:
From the table:
Total number of Girls = 18Total number of students = 30Total number of Girls who are Freshman = 4[tex]\sf Probability \ of \ an \ event \ occurring = \dfrac{Number \ of \ ways \ it \ can \ occur}{Total \ number \ of \ possible \ outcomes}[/tex]
[tex]\sf \implies P(Girl)=\dfrac{18}{30}[/tex]
[tex]\sf \implies P(Freshman \cap Girl)=\dfrac{4}{30}[/tex]
[tex]\begin{aligned}\sf P(Freshman|Girl)& = \sf\dfrac{P(Freshman \cap Girl)}{P(Girl)}\\\\ & = \sf \dfrac{4}{30} \div \dfrac{18}{30}\\\\ & \sf=\dfrac{4}{30} \times \dfrac{30}{18}\\\\ & = \sf \dfrac29\end{aligned}[/tex]
A car is traveling a rate of 120 kilometers per hour. What is the cars rate in mile per hour? How many miles will the car travel in 2 hours? In your computation assume that 1 mike is equal to 1.6 kilometers
Step-by-step explanation:
120 ÷ 1.6 = 75.
so 75 miles per hour.
75 × 2 = 150
therefore 150 miles in 2 hours
Answer:
120 km/ 1 hour
We need to convert kilometers into miles, so given the information below (1 mile = 1.6 km), we divide 120 km by 1.6 km because 1.6 km is equivalent to 1 mile.
120km/1.6km
= 75 miles, so therefore the cars rate will be:-
75 miles/ 1 hour
To see how many miles are in 2 hours, we must multiply the unit rate(75 miles/ 1 hour) by 2.
75 miles/ 1 hour x 2/2 hours
75 x 2
______ = 150/2, so the car will travel 150 miles in 2 hours.
1 x 2
Jackie wants to meet up with her friend who lives 3 miles away and bring him the cookies she baked. she and her friend start moving toward each other at the same time. Jackie's walking speed is 3mph, and her friend's biking speed is 12mph. 4 minutes after Jackie left the house, her mom notices that she forgot the cookies on the kitchen table. She sends Jackie's brother to bring them to Jackie. How fast should Jackie's brother travel to catch up with Jackie before she meets her friend?
Answer:
First, find the time when Jackie and her friend meet.
Using:
[tex]\sf s=ut+\dfrac12at^2[/tex]
where:
s = displacementu = initial velocitya = accelerationt = time (in hours)Jackie and her friend are not accelerating, so a = 0, which means the formula can be reduced to:
[tex]\sf s=ut[/tex]
Create equations
Jackie: [tex]\sf s_1= 3t[/tex]
Friend: [tex]\sf s_2 = 12t[/tex]
As the distance covered by Jackie and her friend at the time they meet equals 3 miles:
[tex]\sf s_1+s_2=3 \ miles[/tex]
[tex]\sf \implies 3t + 12t = 3[/tex]
[tex]\sf \implies 15t = 3[/tex]
[tex]\sf \implies t = 0.2 \ hr[/tex]
[tex]\sf \implies t=0.2 \times 60=12 \ mins[/tex]
Therefore, they meet 12 mins after they both start moving.
3 mph = 3 ÷ 60 = 0.05 miles per min
12 mph = 12 ÷ 60 = 0.2 miles per min
Therefore,
Jackie traveled: 0.05 x 12 = 0.6 miles
Her friend traveled: 0.2 x 12 = 2.4 miles
when they meet.
To find the speed Jackie's brother must walk to catch up with her at the same time she meets her friend, use s = 0.6 miles (as this is the distance Jackie walked) and t = 12 - 4 = 8 mins (since he left 4 mins after Jackie left):
[tex]\sf s=ut[/tex]
[tex]\sf \implies u = \dfrac{s}{t}[/tex]
[tex]\implies \sf u = \dfrac{0.6}{8}[/tex]
[tex]\implies \sf u = 0.075 \ miles \ per \ min[/tex]
[tex]\sf \implies u = 0.075 \times 60 = 4.5 \ mph[/tex]
So Jackie's brother needs to walk faster than 4.5 mph to catch up with Jackie before she meets her friend.
(If he walks at 4.5 mph, he will catch up to Jackie at the instant she meets her friend. If he walks slower than 4.5 mph, he will not catch up with Jackie before she meets her friend).
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Answer:
Hi shelp >//<
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Answer:
Hi shelp >//<
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Step-by-step explanation:
enebe eeeHAHAHA
a. 25
b. 32
c. 27.2
d. 26.4
Answer:
27.2
Step-by-step explanation:
sin 54 = opposite / hyp = 22 / x
sin 54 = 22/x
x = 22/sin54 = 27.2
Find all the common factors of 8 and 12.
A) 2, 4
B) 1, 2, 4
C) 1, 2, 4, 8
Answer:
B is the answer
Step-by-step explanation:
Hope it helps.
Single post math redo question
Answer:
See attached graph
Step-by-step explanation:
We need to convert the polar coordinates into Cartesian coordinates using the rules [tex]x=r\:cos\theta[/tex] and [tex]y=r\:sin\theta[/tex] (see table in attached file).
The converted points will start to resemble a circle with a horizontal pole and a diameter of 4, so by connecting the points, we get our equation [tex]r=4\:cos\theta[/tex].
CHERRY PIE A circular cherry pie has a radius of 6 inches. If the pie is cut into 8 congruent slices, what is
the area of one slice to the nearest hundredth?
6 in.
16.35 in?
14.14 in?
19.72 in2
17.13 in?
170 What is the mZBAC?
Answer:
m<BAC = 85°
Step-by-step explanation:
The measure of an inscribed angle is half the measure of its intercepted arc.
m<BAC = 170°/2
m<BAC = 85°
Suppose we want to choose 7 letters, without replacement, from 12 distinct letters. (a) How many ways can this be done, if the order of the choices is not relevant?
Answer:
3991680 or 792
Step-by-step explanation:
12P7 = 3991680
or
12C7 = 792
P = permutation
C = combination
Sorry, I am not sure which answer is correct or if any of the answers are correct.
Kim is 15 years old. Her father is 4 times as old as her now. In how many years time their total age be 115 years.
Answer:
20 years time
Step-by-step explanation:
I'LL MARK AS THE BRAINLIEST!!!!! a− (10 − a) = 30
Answer:
a=20
Step-by-step explanation:
a-10+a=30
a+a-10=30
2a-10=30
2a=30+10
2a=40
a=20
The average number of potholes per 10 miles of paved U.S. roads is 130. Assume this variable is approximately normally distributed and has a standard deviation of 5. Find the probability that a randomly selected road has more than 142 potholes per 10 miles
Answer:
54.03% probability that a randomly selected road has between 128 and 136 potholes per 10 miles.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
Find the probability that a randomly selected road has between 128 and 136 potholes per 10 miles.
This probability is the pvalue of Z when X = 136 subtracted by the pvalue of Z when X = 128. So
X = 136
has a pvalue of 0.8849.
X = 128
has a pvalue of 0.3446.
0.8849 - 0.3446 = 0.5403
54.03% probability that a randomly selected road has between 128 and 136 potholes per 10 miles.
how do i solve this and what’s there answer?
[tex]\qquad\qquad\huge\underline{{\sf Answer♪}}[/tex]
If the given triangles are congruent, then their corresponding sides are equal as well ~
So, let's use this condition
[tex]\qquad \sf \dashrightarrow \:2y + 5 = 3y + 2[/tex]
[tex]\qquad \sf \dashrightarrow \:3y - 2y = 5 - 2[/tex]
[tex]\qquad \sf \dashrightarrow \:y = 3[/tex]
and
[tex]\qquad \sf \dashrightarrow \:2x + 7 = 15[/tex]
[tex]\qquad \sf \dashrightarrow \:2x = 8[/tex]
[tex]\qquad \sf \dashrightarrow \:x = 4[/tex]