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tester [92]
3 years ago
9

PLEASE HELP !!!!!!!!Solve the system of equations using the linear combination method.

Mathematics
1 answer:
vfiekz [6]3 years ago
6 0

Answer:

The solution would be (5, -2)

Step-by-step explanation:

To use this method, start by multiplying the second equation by -1. Then add the two equations together.

9x + 5y = 35

-2x - 5y = 0

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

7x = 35

x = 5

Now that we have the value of x, use it to solve either equation for y.

2x + 5y = 0

2(5) + 5y = 0

10 + 5y = 0

5y = -10

y = -2

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Step-by-step explanation:

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3 years ago
If mints sell three for<br>25°, how many can 1<br>buy with a $20 bill?​
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3 years ago
Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (
tekilochka [14]

Answer:

Radius of convergence of power series is \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n        n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n       n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

Power series centered at x = a is:

\sum_{n=1}^{\infty}c_{n}(x-a)^{n}

\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\

a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]

Applying the ratio test:

\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}

\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

Applying n → ∞

\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}

The numerator as well denominator of \frac{a_{n}}{a_{n+1}} are polynomials of fifth degree with leading coefficients:

(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}

4 0
3 years ago
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