Answer:
.
Step-by-step explanation:
The equation is
y = 200+50x
This is not a proportional relationship because it does not go through the origin
If x=0, y does not equal 0
Answer:
Alternate-exterior angles theorem.
Step-by-step explanation:
Two parallel lines are cut by a transversal, and if there are a pair of congruent angles that are outside of the parallel lines, and on opposite sides of the transversal, you will have the alternate-exterior angles theorem.
You can think of x && y || z as equivalent to: int func (int x, int y, int z) { if (x) { if (y) { return true; } } if (z) { return true; } return false; } Since both x and y are fixed to be non-zero values the first return statement is always hit. thats what i think
To solve this, first we calculate all the total cost of
the items that is:
total cost = $14.96 + $19.87 + $5.37
total cost = $40.20
So we see that the actual cost is similar with the
estimate so his calculation is reasonable.
Looks like your function is
![f(x)=\dfrac x{6x^2+1}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%20x%7B6x%5E2%2B1%7D)
Rewrite it as
![f(x)=\dfrac x{1-(-6x^2)}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%20x%7B1-%28-6x%5E2%29%7D)
Recall that for
, we have
![\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n](https://tex.z-dn.net/?f=%5Cdfrac1%7B1-x%7D%3D%5Cdisplaystyle%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20x%5En)
If we replace
with
, we get
![f(x)=\displaystyle x\sum_{n=0}^\infty\frac(-6x^2)^n=\sum_{n=0}^\infty (-6)^n x^{2n+1}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdisplaystyle%20x%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%28-6x%5E2%29%5En%3D%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20%28-6%29%5En%20x%5E%7B2n%2B1%7D)
By the ratio test, the series converges if
![\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cleft%7C%5Cfrac%7B%28-6%29%5E%7Bn%2B1%7D%20x%5E%7B2%28n%2B1%29%2B1%7D%7D%7B%28-6%29%5En%20x%5E%7B2n%2B1%7D%7D%5Cright%7C%3D6%7Cx%5E2%7C%5Clim_%7Bn%5Cto%5Cinfty%7D1%3D6%7Cx%7C%5E2%3C1)
Solving for
gives the interval of convergence,
![|x|^2](https://tex.z-dn.net/?f=%7Cx%7C%5E2%3C%5Cdfrac16%5Cimplies%7Cx%7C%3C%5Cdfrac1%7B%5Csqrt6%7D%5Cimplies%20-%5Cdfrac1%7B%5Csqrt%206%7D%3Cx%3C%5Cdfrac1%7B%5Csqrt%206%7D)
We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.