Answer:
The length of the missing side is 5 m
Step-by-step explanation:
In order to find this, use the perimeter formula.
P = 2L + 2W
Now input 11 as the L and 32 as P
32 = 2(11) + 2W
32 = 22 + 2W
10 = 2W
5 = W
let's change some the 0.1 to say 1/10, just the fraction version of it.
![\bf \lim\limits_{x\to \left( \frac{1}{10} \right)^-}~\cfrac{10x-1}{|10x^3-x^2|}\implies \lim\limits_{x\to \left( \frac{1}{10} \right)^-}~\cfrac{10(-x)-1}{10(-x)^3-(-x)^2}](https://tex.z-dn.net/?f=%5Cbf%20%5Clim%5Climits_%7Bx%5Cto%20%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29%5E-%7D~%5Ccfrac%7B10x-1%7D%7B%7C10x%5E3-x%5E2%7C%7D%5Cimplies%20%5Clim%5Climits_%7Bx%5Cto%20%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29%5E-%7D~%5Ccfrac%7B10%28-x%29-1%7D%7B10%28-x%29%5E3-%28-x%29%5E2%7D)
![\bf \cfrac{-10x-1}{-10x^3-x^2}\implies \cfrac{-10\left( \frac{1}{10} \right)-1}{-10\left( \frac{1}{10} \right)^3-\left( \frac{1}{10} \right)^2}\implies \cfrac{-1-1}{-\frac{1}{100}-\frac{1}{100}}\implies \cfrac{-2}{\frac{-2}{100}} \\\\\\ \cfrac{~~\begin{matrix} -2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{1}\cdot \cfrac{100}{~~\begin{matrix} -2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies 100](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B-10x-1%7D%7B-10x%5E3-x%5E2%7D%5Cimplies%20%5Ccfrac%7B-10%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29-1%7D%7B-10%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29%5E3-%5Cleft%28%20%5Cfrac%7B1%7D%7B10%7D%20%5Cright%29%5E2%7D%5Cimplies%20%5Ccfrac%7B-1-1%7D%7B-%5Cfrac%7B1%7D%7B100%7D-%5Cfrac%7B1%7D%7B100%7D%7D%5Cimplies%20%5Ccfrac%7B-2%7D%7B%5Cfrac%7B-2%7D%7B100%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B~~%5Cbegin%7Bmatrix%7D%20-2%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%7B1%7D%5Ccdot%20%5Ccfrac%7B100%7D%7B~~%5Cbegin%7Bmatrix%7D%20-2%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%7D%5Cimplies%20100)
when checking an absolute value expression, we do the one-sided limits, since an absolute value expression is in effect a piecewise function with ± versions, so for the limit from the left we check the negative version.
The set of all values for which the function is defined
Answer:
<h2>
a₁₀₀ = 385</h2>
Step-by-step explanation:
The nth term of arithmetic sequence is ![a_n=a_1+d(n-1)](https://tex.z-dn.net/?f=a_n%3Da_1%2Bd%28n-1%29)
So:
![a_9=a_1+d(9-1)\\\\37=a_1+8d\\\\a_1=37-8d\\\\\\a_{16}=a_1+d(16-1)\\\\65=a_1+15d\\\\65=37-8d+15d\\\\28=7d\\\\d=4\\\\a_1=37-8\cdot4=5\\\\\\a_{100}=a_1+d(100-1)=5+4\cdot95=385](https://tex.z-dn.net/?f=a_9%3Da_1%2Bd%289-1%29%5C%5C%5C%5C37%3Da_1%2B8d%5C%5C%5C%5Ca_1%3D37-8d%5C%5C%5C%5C%5C%5Ca_%7B16%7D%3Da_1%2Bd%2816-1%29%5C%5C%5C%5C65%3Da_1%2B15d%5C%5C%5C%5C65%3D37-8d%2B15d%5C%5C%5C%5C28%3D7d%5C%5C%5C%5Cd%3D4%5C%5C%5C%5Ca_1%3D37-8%5Ccdot4%3D5%5C%5C%5C%5C%5C%5Ca_%7B100%7D%3Da_1%2Bd%28100-1%29%3D5%2B4%5Ccdot95%3D385)
![\sf\dfrac{3}{11}-(-\dfrac{6}{11})](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7B3%7D%7B11%7D-%28-%5Cdfrac%7B6%7D%7B11%7D%29)
Two negatives equals a positive:
![\sf\dfrac{3}{11}+\dfrac{6}{11}](https://tex.z-dn.net/?f=%5Csf%5Cdfrac%7B3%7D%7B11%7D%2B%5Cdfrac%7B6%7D%7B11%7D)
Add the numerators and keep the denominator: