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Eva8 [605]
3 years ago
10

find to consecutivephone number such that five times the smaller number plus 3 times the greater number make 59 hint let the num

ber be x and x + 1​
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
7 0

Answer:

x = 7 is the answer for the smaller number

7 + 1 = 8 is the answer for the larger number

Step-by-step explanation:

LET x and  x+1 be the two numbers here, as your hint suggests.

5x + 3(x+1) = 59

5x + 3x+3 = 59

8x = 59-3

8x = 56

/8      /8

--------------------------

x = 56/8 ---> x = 7

x = 7 is the answer for the smaller number

7 + 1 = 8 is the answer for the larger number

Check:

5 * 7 + 3 * 8 = 59

35 + 24 = 59

59 = 59 CORRECT

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Solve using Fourier series.
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\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
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Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

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\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
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