Ten elevenths as a decimal would be 0.90909 because you would divide ten by eleven and then you would get 0.9090.
The answer is 18. The two sides are the same length, so you set the two sides equal to each other and solve for x. After finding x, you plug in the number into the base expression and you get the base length.
Answer:
.
Step-by-step explanation:
To find
.
First, calculate the corresponding indefinite integral:
Integrate term by term:
![\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u}} =\int{2 d u} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}](https://tex.z-dn.net/?f=%5Cint%7B%5Cleft%28-%20%5Cfrac%7B3%20u%5E%7B9%7D%7D%7B2%7D%20%2B%20%5Cfrac%7B2%20u%5E%7B4%7D%7D%7B5%7D%20%2B%202%5Cright%29d%20u%7D%7D%20%3D%5Cint%7B2%20d%20u%7D%20%2B%20%5Cint%7B%5Cfrac%7B2%20u%5E%7B4%7D%7D%7B5%7D%20d%20u%7D%20-%20%5Cint%7B%5Cfrac%7B3%20u%5E%7B9%7D%7D%7B2%7D%20d%20u%7D)
Apply the constant rule ![\int c\, du = c u](https://tex.z-dn.net/?f=%5Cint%20c%5C%2C%20du%20%3D%20c%20u)
![\int{2 d u}} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u} = {\left(2 u\right)} + \int{\frac{2 u^{4}}{5} d u} - \int{\frac{3 u^{9}}{2} d u}](https://tex.z-dn.net/?f=%5Cint%7B2%20d%20u%7D%7D%20%2B%20%5Cint%7B%5Cfrac%7B2%20u%5E%7B4%7D%7D%7B5%7D%20d%20u%7D%20-%20%5Cint%7B%5Cfrac%7B3%20u%5E%7B9%7D%7D%7B2%7D%20d%20u%7D%20%3D%20%7B%5Cleft%282%20u%5Cright%29%7D%20%2B%20%5Cint%7B%5Cfrac%7B2%20u%5E%7B4%7D%7D%7B5%7D%20d%20u%7D%20-%20%5Cint%7B%5Cfrac%7B3%20u%5E%7B9%7D%7D%7B2%7D%20d%20u%7D)
Apply the constant multiple rule ![\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du](https://tex.z-dn.net/?f=%5Cint%20c%20f%7B%5Cleft%28u%20%5Cright%29%7D%5C%2C%20du%20%3D%20c%20%5Cint%20f%7B%5Cleft%28u%20%5Cright%29%7D%5C%2C%20du)
![2 u - {\int{\frac{3 u^{9}}{2} d u}} + \int{\frac{2 u^{4}}{5} d u} = 2 u - {\left(\frac{3}{2} \int{u^{9} d u}\right)} + \left(\frac{2}{5} \int{u^{4} d u}\right)](https://tex.z-dn.net/?f=2%20u%20-%20%7B%5Cint%7B%5Cfrac%7B3%20u%5E%7B9%7D%7D%7B2%7D%20d%20u%7D%7D%20%2B%20%5Cint%7B%5Cfrac%7B2%20u%5E%7B4%7D%7D%7B5%7D%20d%20u%7D%20%3D%202%20u%20-%20%7B%5Cleft%28%5Cfrac%7B3%7D%7B2%7D%20%5Cint%7Bu%5E%7B9%7D%20d%20u%7D%5Cright%29%7D%20%2B%20%5Cleft%28%5Cfrac%7B2%7D%7B5%7D%20%5Cint%7Bu%5E%7B4%7D%20d%20u%7D%5Cright%29)
Apply the power rule ![\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}](https://tex.z-dn.net/?f=%5Cint%20u%5E%7Bn%7D%5C%2C%20du%20%3D%20%5Cfrac%7Bu%5E%7Bn%20%2B%201%7D%7D%7Bn%20%2B%201%7D)
![2 u - \frac{3}{2} {\int{u^{9} d u}} + \frac{2}{5} {\int{u^{4} d u}}=2 u - \frac{3}{2} {\frac{u^{1 + 9}}{1 + 9}}+ \frac{2}{5}{\frac{u^{1 + 4}}{1 + 4}}](https://tex.z-dn.net/?f=2%20u%20-%20%5Cfrac%7B3%7D%7B2%7D%20%7B%5Cint%7Bu%5E%7B9%7D%20d%20u%7D%7D%20%2B%20%5Cfrac%7B2%7D%7B5%7D%20%7B%5Cint%7Bu%5E%7B4%7D%20d%20u%7D%7D%3D2%20u%20-%20%5Cfrac%7B3%7D%7B2%7D%20%7B%5Cfrac%7Bu%5E%7B1%20%2B%209%7D%7D%7B1%20%2B%209%7D%7D%2B%20%5Cfrac%7B2%7D%7B5%7D%7B%5Cfrac%7Bu%5E%7B1%20%2B%204%7D%7D%7B1%20%2B%204%7D%7D)
Therefore,
![\int{\left(- \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2\right)d u} = - \frac{3 u^{10}}{20} + \frac{2 u^{5}}{25} + 2 u = \frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)](https://tex.z-dn.net/?f=%5Cint%7B%5Cleft%28-%20%5Cfrac%7B3%20u%5E%7B9%7D%7D%7B2%7D%20%2B%20%5Cfrac%7B2%20u%5E%7B4%7D%7D%7B5%7D%20%2B%202%5Cright%29d%20u%7D%20%3D%20-%20%5Cfrac%7B3%20u%5E%7B10%7D%7D%7B20%7D%20%2B%20%5Cfrac%7B2%20u%5E%7B5%7D%7D%7B25%7D%20%2B%202%20u%20%3D%20%5Cfrac%7Bu%7D%7B100%7D%20%5Cleft%28-%2015%20u%5E%7B9%7D%20%2B%208%20u%5E%7B4%7D%20%2B%20200%5Cright%29)
According to the Fundamental Theorem of Calculus,
, so just evaluate the integral at the endpoints, and that's the answer.
![\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}=\frac{193}{100}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bu%7D%7B100%7D%20%5Cleft%28-%2015%20u%5E%7B9%7D%20%2B%208%20u%5E%7B4%7D%20%2B%20200%5Cright%29%5Cright%29%7C_%7B%5Cleft%28u%3D1%5Cright%29%7D%3D%5Cfrac%7B193%7D%7B100%7D)
![\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=0](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7Bu%7D%7B100%7D%20%5Cleft%28-%2015%20u%5E%7B9%7D%20%2B%208%20u%5E%7B4%7D%20%2B%20200%5Cright%29%5Cright%29%7C_%7B%5Cleft%28u%3D0%5Cright%29%7D%3D0)
![\int_{0}^{1}\left( - \frac{3 u^{9}}{2} + \frac{2 u^{4}}{5} + 2 \right)du=\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=1\right)}-\left(\frac{u}{100} \left(- 15 u^{9} + 8 u^{4} + 200\right)\right)|_{\left(u=0\right)}=\frac{193}{100}](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B1%7D%5Cleft%28%20-%20%5Cfrac%7B3%20u%5E%7B9%7D%7D%7B2%7D%20%2B%20%5Cfrac%7B2%20u%5E%7B4%7D%7D%7B5%7D%20%2B%202%20%5Cright%29du%3D%5Cleft%28%5Cfrac%7Bu%7D%7B100%7D%20%5Cleft%28-%2015%20u%5E%7B9%7D%20%2B%208%20u%5E%7B4%7D%20%2B%20200%5Cright%29%5Cright%29%7C_%7B%5Cleft%28u%3D1%5Cright%29%7D-%5Cleft%28%5Cfrac%7Bu%7D%7B100%7D%20%5Cleft%28-%2015%20u%5E%7B9%7D%20%2B%208%20u%5E%7B4%7D%20%2B%20200%5Cright%29%5Cright%29%7C_%7B%5Cleft%28u%3D0%5Cright%29%7D%3D%5Cfrac%7B193%7D%7B100%7D)
we know that 9*9 = 81, so let's start there.
0.9 * 0.9
one amount has 1 decimal, the second amount has another decimal, so in the product, we must have two decimals, let's do the multiplication without the dot.
09 * 09 = 81
but we must have two decimals, so that becomes 0.81.
0.09 * 0.9
here we do the same, the multiplication without the dot, and then we add as many decimals, in this case is 3 decimals, first term has 2 and the second term has 1 decimal, so the product must have 3 decimals.
009 * 09 = 81
getting those 3 decimals.......... 0.081.
0.009 * 0.9
here we do the same
0009 * 09 = 81
the product needs 4 decimals, so................ 0.0081.
0.009 * 0.09
we do the same here
0009 * 009 = 81
since the product needs 5 decimals................. 0.00081.