Answer:
Congrats?
Step-by-step explanation:
I don´t understand the question
Answer:
x + 2 < 6
x < 4
d represents the inequality
Answer:
0.629 or 0.795
Step-by-step explanation:
Answer:
<h2> StartFraction 7 over 10 EndFraction x + 2 and one-half y + 6</h2>
Step-by-step explanation:
Given the expression ![2(\frac{3x}{5}+2\frac{3y}{4}-\frac{x}{4}-1 \frac{1}{2}y+3)](https://tex.z-dn.net/?f=2%28%5Cfrac%7B3x%7D%7B5%7D%2B2%5Cfrac%7B3y%7D%7B4%7D-%5Cfrac%7Bx%7D%7B4%7D-1%20%5Cfrac%7B1%7D%7B2%7Dy%2B3%29)
To simplify the expression, we need to first collect the like terms of the functions in parentheses as shown;
![= 2(\frac{3x}{5}-\frac{x}{4}-1 \frac{1}{2}y+2\frac{3}{4}y+3)\\= 2(\frac{3x}{5}-\frac{x}{4}- \frac{3}{2}y+\frac{11}{4}y+3)\\](https://tex.z-dn.net/?f=%3D%202%28%5Cfrac%7B3x%7D%7B5%7D-%5Cfrac%7Bx%7D%7B4%7D-1%20%5Cfrac%7B1%7D%7B2%7Dy%2B2%5Cfrac%7B3%7D%7B4%7Dy%2B3%29%5C%5C%3D%202%28%5Cfrac%7B3x%7D%7B5%7D-%5Cfrac%7Bx%7D%7B4%7D-%20%5Cfrac%7B3%7D%7B2%7Dy%2B%5Cfrac%7B11%7D%7B4%7Dy%2B3%29%5C%5C)
Then we find the LCM of the resulting function
![= 2(\frac{3x}{5}-\frac{x}{4}- \frac{3}{2}y+\frac{11}{4}y+3)\\= 2(\frac{12x-5x}{20} - (\frac{6y-11y}{4})+3)\\= 2(\frac{7x}{20}- (\frac{-5y}{4})+3 )\\= 2(\frac{7x}{20}+ \frac{5y}{4}+3 )\\= \frac{7x}{10} + \frac{5y}{2} +6\\= \frac{7x}{10} + 2\frac{1}{2}y+6\\](https://tex.z-dn.net/?f=%3D%202%28%5Cfrac%7B3x%7D%7B5%7D-%5Cfrac%7Bx%7D%7B4%7D-%20%5Cfrac%7B3%7D%7B2%7Dy%2B%5Cfrac%7B11%7D%7B4%7Dy%2B3%29%5C%5C%3D%202%28%5Cfrac%7B12x-5x%7D%7B20%7D%20-%20%28%5Cfrac%7B6y-11y%7D%7B4%7D%29%2B3%29%5C%5C%3D%202%28%5Cfrac%7B7x%7D%7B20%7D-%20%28%5Cfrac%7B-5y%7D%7B4%7D%29%2B3%20%29%5C%5C%3D%202%28%5Cfrac%7B7x%7D%7B20%7D%2B%20%5Cfrac%7B5y%7D%7B4%7D%2B3%20%29%5C%5C%3D%20%5Cfrac%7B7x%7D%7B10%7D%20%2B%20%5Cfrac%7B5y%7D%7B2%7D%20%2B6%5C%5C%3D%20%20%5Cfrac%7B7x%7D%7B10%7D%20%2B%202%5Cfrac%7B1%7D%7B2%7Dy%2B6%5C%5C)
The final expression gives the required answer
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)