If two similar triangles have side lengths 3cm and 7cm then we can determine that the scale factor for the lengths is: 7/3
Now, we have to work out the area scale factor !
This is simple, since area is given in cm² then all we have to do is square our original scale factor!
So,
7/3² = 49/9
So, we can say that the ratio of the area would be:
x : 49/9x
Hope this helps! <3
When a set of objects are chosen from a larger set in which the order of the object doesn't matter, we call this combination. Otherwise if the order of selection would matter this would be called a permutation.
Probability is the likelihood of selecting a particular specified object in from a specified group of objects.
Therefore our answer is B.
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:
It's a probability problem to find the odds of picking a green or red
shirt out of the 10 shirts on Thursday, Friday and Saturday since you
have randomly already know you have picked a blue shirt on the other
days. Not sure if you have this as a multiple question problem as you didn't list any possible answers (A. 7/20, B. 5/47, C. 2/5, D. 4/125) to the question. A, B, C, D being like 7 chances out of 20, 5 chances out 47, 2 chances out of chances 5 or 4 chances out of 125 (example answers only).
Probability = Number favorable outcomes / total number of outcomes
Use the identity P(A ∪ B) = P(A)+P(B)-P(A ∩ B)
P(A)=0.50
P(B)=0.60
P(A ∪ B) = 0.30
=>
P(A ∪ B) = P(A)+P(B)-P(A ∩ B)
=(0.50+0.60)-0.30
=0.80
