$\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\[-0.35em] ~\dotfill$

$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \cfrac{\textit{figure A}}{\textit{figure B}}\qquad \qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{512}}{\sqrt[3]{343}}\qquad \begin{cases} 512=&2^9\\ &2^{3\cdot 3}\\ &(2^3)^3\\ 343=&7^3 \end{cases}\implies \cfrac{s}{s}=\cfrac{\sqrt[3]{(2^3)^3}}{\sqrt[3]{7^3}} \\\\\\ \cfrac{s}{s}=\cfrac{2^3}{7}\implies \cfrac{s}{s}=\cfrac{8}{7}\implies s:s = 8:7\impliedby \textit{ratio of the }\stackrel{sides~and}{perimeters}$

5 0

3 years ago

Find the other endpoint of a line segment which one endpoint at (-3, 5) and whose midpoint is (5, 2)

Feliz [49]

So to find the other endpoint you need to see how far the first endpoint is to the midpoint. In X's place we have a total distance of
5- (-3)= 8 place away from one another so
5+8= 13

while in the y's case
5-2=3 place away from one another so
2-3= -1

so your order pair is
(13,-1)

7 0

3 years ago

Read 2 more answers

I need the formula can you please help me

Ahat [919]

Multiply the diagonals together and then divide by two

4 x 6 / 2

so the answer would be 12yd

5 0

3 years ago

Read 2 more answers

I also need help with this quick!

Jlenok [28]

Answer: