![\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\ \rule{31em}{0.25pt}\\\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bratio%20relations%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bccccllll%7D%20%26%5Cstackrel%7Bratio~of~the%7D%7BSides%7D%26%5Cstackrel%7Bratio~of~the%7D%7BAreas%7D%26%5Cstackrel%7Bratio~of~the%7D%7BVolumes%7D%5C%5C%20%26-----%26-----%26-----%5C%5C%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%26%5Ccfrac%7Bs%7D%7Bs%7D%26%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%26%5Ccfrac%7Bs%5E3%7D%7Bs%5E3%7D%20%5Cend%7Barray%7D%5C%5C%5C%5C%20%5Crule%7B31em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%20)
![\bf \rule{31em}{0.25pt}\\\\ \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{98}}{\sqrt{162}}~~ \begin{cases} 98=2\cdot 7\cdot 7\\ \qquad 2\cdot 7^2\\ 162=2\cdot 9\cdot 9\\ \qquad 2\cdot 9^2 \end{cases}\implies \cfrac{s}{s}=\cfrac{\sqrt{2\cdot 7^2}}{\sqrt{2\cdot 9^2}} \\[2em] \cfrac{s}{s}=\cfrac{7\sqrt{2}}{9\sqrt{2}}\implies \cfrac{s}{s}=\cfrac{7}{9}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Crule%7B31em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Ccfrac%7Bsmaller%7D%7Blarger%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7B98%7D%7D%7B%5Csqrt%7B162%7D%7D~~%20%5Cbegin%7Bcases%7D%2098%3D2%5Ccdot%207%5Ccdot%207%5C%5C%20%5Cqquad%202%5Ccdot%207%5E2%5C%5C%20162%3D2%5Ccdot%209%5Ccdot%209%5C%5C%20%5Cqquad%202%5Ccdot%209%5E2%20%5Cend%7Bcases%7D%5Cimplies%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7B2%5Ccdot%207%5E2%7D%7D%7B%5Csqrt%7B2%5Ccdot%209%5E2%7D%7D%20%5C%5C%5B2em%5D%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B7%5Csqrt%7B2%7D%7D%7B9%5Csqrt%7B2%7D%7D%5Cimplies%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B7%7D%7B9%7D%20)
bearing in mind that the ratio of the sides, is the same as the ratio of the perimeters.
The problem given above is simply a conversion problem. <span>We
are asked to convert from units of acres to units hectares. To do
this, we need a conversion factor which would relate the different units
involved. We either multiply or divide this certain value to the original
measurement depending on what is asked. For this case, we use the given values to determine the relation. We do as follows:
1 hectare = 100 m x 100 m = 10000 m^2
1 hectare = 10000 m^2 ( 1 ft / 0.3048 m )^2 = 107639.10 ft^2
107639.10 ft^2 ( 1 acre / 43600 ft^2 ) = 2.47 acre
Thus, 1 hectare = 2.47 acres
12.0 acres ( 1 hectare / 2.47 acres ) = 4.86 hectares
</span><span>
</span>
Answer:
the answer is 5/6 also knows as answer #3
Step-by-step explanation:
because there is 6 sides #s 1-6
Answer:
This is not possible with 2 equal symbols
Answer:
<h2>g = - 3</h2>
Step-by-step explanation:
5+4g+8=1
To solve the equation first group like terms.
Send the constants to the right side of the equation
That's
4g = 1 - 5 - 8
Simplify
4g = - 12
Divide both sides by 4
We have the final answer as
<h3>g = - 3</h3>
Hope this helps you
