It would be 16.5 inches long.
First you have to divide 11 by 8 (1.375) to get how long would the letter be if it was 1 inch wide. Then multiply 1.375 by 12 which is 16.5.
Answer:
graph these
y = 2x + 2
y = 2x - 3
Step-by-step explanation:
2x - y = -2
2x - y = 3
convert to y intercept form
first
2x - y = -2
-y = -2x - 2
y = 2x + 2
second
2x - y = 3
-y = -2x + 3
y = 2x - 3
you are left with
y = 2x + 2
y = 2x - 3
i cannot graph these for you, but i assume you know how, there is no solution because the lines are parallel
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:
r = i + j + (-2/3)(3i - j)
Step-by-step explanation:
Vector Equation of a line - r = a + kb ; where r is the resultant vector of adding vector a and vector b and k is a constant
if a = i + j ; b = t(3i - j) and r = -i +s(j)
for this to be true all the vector components must be equal
summing i 's
i + 3ti = -i; then t = -2/3
j - tj = sj; then s = 1-t; substitue t; s=1+2/3 = 5/3
so r = i + j + (-2/3)(3i - j) which will symplify to -i + 5/3j
<span>solve the equation ax – c = bx + d for x:
1) Group the x terms together on the left: ax - bx - c = d
2) Group the constant terms together: ax - bx = c + d
3) factor out x: x(a - b) = c + d
4) Divide both sides of the equation by (a - b) to obtain a formula for x:
c+d
</span> x(a - b) = c + d => x = ---------
a-b
This shows that the given equation CAN be solved for x, but there is a restriction: a must NOT equal b, because if a-b = 0, we'd have division by zero (which is not defined).
Where is Victoria's solution? Please share it if you want to discuss this problem further. Thank you.
