It will be 5 cm and use 3.14 to multiply
Answer: 15309
hope this helps, the nth term calulation is (7/3) times 3^n
Answer:
![e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...](https://tex.z-dn.net/?f=e%5E%7B4x%7D%3De%5E4%2B4e%5E4%28x-1%29%2B8e%5E4%28x-1%29%5E2%2B...)
![\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n](https://tex.z-dn.net/?f=%5Cdisplaystyle%20e%5E%7B4x%7D%3D%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%20%5Cdfrac%7B4%5Ene%5E4%7D%7Bn%21%7D%28x-1%29%5En)
Step-by-step explanation:
<u>Taylor series</u> expansions of f(x) at the point x = a
![\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...](https://tex.z-dn.net/?f=%5Ctext%7Bf%7D%28x%29%3D%5Ctext%7Bf%7D%28a%29%2B%5Ctext%7Bf%7D%5C%3A%27%28a%29%28x-a%29%2B%5Cdfrac%7B%5Ctext%7Bf%7D%5C%3A%27%27%28a%29%7D%7B2%21%7D%28x-a%29%5E2%2B%5Cdfrac%7B%5Ctext%7Bf%7D%5C%3A%27%27%27%28a%29%7D%7B3%21%7D%28x-a%29%5E3%2B...%2B%5Cdfrac%7B%5Ctext%7Bf%7D%5C%3A%5E%7B%28r%29%7D%28a%29%7D%7Br%21%7D%28x-a%29%5Er%2B...)
This expansion is valid only if
exists and is finite for all
, and for values of x for which the infinite series converges.
![\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1](https://tex.z-dn.net/?f=%5Ctextsf%7BLet%20%7D%5Ctext%7Bf%7D%28x%29%3De%5E%7B4x%7D%20%5Ctextsf%7B%20and%20%7Da%3D1)
![\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...](https://tex.z-dn.net/?f=%5Ctext%7Bf%7D%28x%29%3D%5Ctext%7Bf%7D%281%29%2B%5Ctext%7Bf%7D%5C%3A%27%281%29%28x-1%29%2B%5Cdfrac%7B%5Ctext%7Bf%7D%5C%3A%27%27%281%29%7D%7B2%21%7D%28x-1%29%5E2%2B...)
![\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbegin%7Bminipage%7D%7B5.5%20cm%7D%5Cunderline%7BDifferentiating%20%24e%5E%7Bf%28x%29%7D%24%7D%5C%5C%5C%5CIf%20%20%24y%3De%5E%7Bf%28x%29%7D%24%2C%20then%20%24%5Cdfrac%7B%5Ctext%7Bd%7Dy%7D%7B%5Ctext%7Bd%7Dx%7D%3Df%5C%3A%27%28x%29e%5E%7Bf%28x%29%7D%24%5C%5C%5Cend%7Bminipage%7D%7D)
![\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4](https://tex.z-dn.net/?f=%5Ctext%7Bf%7D%28x%29%3De%5E%7B4x%7D%20%5Cimplies%20%5Ctext%7Bf%7D%281%29%3De%5E4)
![\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4](https://tex.z-dn.net/?f=%5Ctext%7Bf%7D%5C%3A%27%28x%29%3D4e%5E%7B4x%7D%20%5Cimplies%20%5Ctext%7Bf%7D%5C%3A%27%281%29%3D4e%5E4)
![\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4](https://tex.z-dn.net/?f=%5Ctext%7Bf%7D%5C%3A%27%27%28x%29%3D16e%5E%7B4x%7D%20%5Cimplies%20%5Ctext%7Bf%7D%5C%3A%27%27%281%29%3D16e%5E4)
Substituting the values in the series expansion gives:
![e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...](https://tex.z-dn.net/?f=e%5E%7B4x%7D%3De%5E4%2B4e%5E4%28x-1%29%2B%5Cdfrac%7B16e%5E4%7D%7B2%7D%28x-1%29%5E2%2B...)
Factoring out e⁴:
![e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]](https://tex.z-dn.net/?f=e%5E%7B4x%7D%3De%5E4%5Cleft%5B1%2B4%28x-1%29%2B8%7D%28x-1%29%5E2%2B...%5Cright%5D)
<u>Taylor Series summation notation</u>:
![\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7Bf%7D%28x%29%3D%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%20%5Cdfrac%7B%5Ctext%7Bf%7D%5C%3A%5E%7B%28n%29%7D%28a%29%7D%7Bn%21%7D%28x-a%29%5En)
Therefore:
![\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n](https://tex.z-dn.net/?f=%5Cdisplaystyle%20e%5E%7B4x%7D%3D%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%20%5Cdfrac%7B4%5Ene%5E4%7D%7Bn%21%7D%28x-1%29%5En)
Answer:
There are 2^3 = 8 possible outcomes after tossing a fair coin fairly 3 times. The 8 possible outcomes are: TTT, HTT, THT, TTH, HHT, HTH, THH, HHH. Exactly 2 of 8 possible outcomes result in 3 of the same faces showing up.
Step-by-step explanation:
This question is super simple when you sit down and think about what it's asking. What is the cost of the whole paving lot?
We know that the paving lot is 90 ft by 75 ft, we need to start by finding the area of the whole paving lot.
Area of a rectangle is Length multiplied by Width.
90 × 75 = 6,750
The area of the paving lot is 6,750 feet squared.
Now we need to find the cost of the paving lot, we know that one square yard costs $3.50 to pave.
We can solve this by first taking the step to finding how many yards are in the paving lot.
A yard is 3 feet so we'll divide the area of the paving lot by 3.
6,750 ÷ 3 = 2,250
With the paving lot having 2,250 square yards we can now multiply this by the $3.5 cost per square yard.
2,250 × 3.5 = 7,875
It should cost $7,875 dollars to pave the entire lot.