Answer:
The roots of the equation 2m²+3=m are non-real roots.
Step-by-step explanation:
Given equation:
2m²+3=m
2m²-m+3=0
Here, from the equation we can obtain the following values:
a = 2, b = -1, c = 3
Discriminant of an equation is given as:
D = b²-4ac
= (-1)²-4(2)(3)
= 1 - 21
= -20
Discriminant can tell what kinds of roots the equation have.
In our case, the discriminant is less than 0.
When D < 0, the roots of the equation are complex conjugates.(non-real)
Answer:
Radius of convergence of power series is 
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}\\](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%288x%2B6%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D2%5E%7Bn%7D%284x%2B3%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%28x%2B%5Cfrac%7B3%7D%7B4%7D%29%5E%7Bn%7D%5C%5C)
Power series centered at x = a is:
![\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}\\](https://tex.z-dn.net/?f=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%288x%2B6%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B2%5E%7Bn%7D%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D2%5E%7Bn%7D%284x%2B3%29%5E%7Bn%7D%5C%5C%5C%5C%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5B%5Cfrac%7B8%5E%7Bn%7D4%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%28x%2B%5Cfrac%7B3%7D%7B4%7D%29%5E%7Bn%7D%5C%5C)
![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)!!}]](https://tex.z-dn.net/?f=a_%7Bn%7D%3D%5B%5Cfrac%7B8%5E%7Bn%7D4%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%5C%5C%5C%5Ca_%7Bn%2B1%7D%3D%5B%5Cfrac%7B8%5E%7Bn%2B1%7D4%5E%7Bn%2B1%7Dn%21%283%28n%2B1%29%2B3%29%21%282%28n%2B1%29%29%21%21%7D%7B%5B%28n%2B1%2B9%29%21%5D%5E%7B3%7D%284%28n%2B1%29%2B3%29%21%21%7D%5D%5C%5C%5C%5Ca_%7Bn%2B1%7D%3D%5B%5Cfrac%7B8%5E%7Bn%2B1%7D4%5E%7Bn%2B1%7D%28n%2B1%29%21%283n%2B6%29%21%282n%2B2%29%21%21%7D%7B%5B%28n%2B10%29%21%5D%5E%7B3%7D%284n%2B7%29%21%21%7D%5D)
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)!!}]}](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7Bn%7D%7D%7Ba_%7Bn%2B1%7D%7D%3D%5Cfrac%7B%5B%5Cfrac%7B32%5E%7Bn%7Dn%21%283n%2B3%29%21%282n%29%21%21%7D%7B%5B%28n%2B9%29%21%5D%5E%7B3%7D%284n%2B3%29%21%21%7D%5D%7D%7B%5B%5Cfrac%7B32%5E%7Bn%2B1%7D%28n%2B1%29%21%283n%2B6%29%21%282n%2B2%29%21%21%7D%7B%5B%28n%2B10%29%21%5D%5E%7B3%7D%284n%2B7%29%21%21%7D%5D%7D)
Applying n → ∞
The numerator as well denominator of 
are polynomials of fifth degree with leading coefficients:
Answer:
Hey mate here's your answer.....
Step-by-step explanation:
Mathematics is a subject where you study more about numbers and daily life applications.
hope its helps you,
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Answer:
360
Step-by-step explanation:
There are 60 minutes in 1 hour, so 2 hours = 120 minutes.
Multiply 120 by 3 and you get 360.
They would burn 360 calories doing 2 hours of bowling.
