Hi can you please help me<br>i cant answer this question.

The height of one square pyramid is 12 m. A similar pyramid has a height of 6 m. The volume of the larger pyramid is 400 m3. Det

tatiyna

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\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}}\\\\
-------------------------------\\\\
$

$\bf \cfrac{smaller}{larger}\qquad \cfrac{\stackrel{height}{6}}{\stackrel{height}{12}}\implies \cfrac{1}{2}\impliedby \textit{scale factor of the pyramids}\\\\
-------------------------------\\\\
\cfrac{smaller}{larger}\qquad \cfrac{h}{h}=\cfrac{\sqrt{base}}{\sqrt{base}}\implies \cfrac{1}{2}=\sqrt{\cfrac{base}{base}}\implies \left( \cfrac{1}{2} \right)^2=\cfrac{base}{base}
\\\\\\
\cfrac{1^2}{2^2}=\cfrac{base}{base}\implies \cfrac{1}{4}=\cfrac{base}{base}\\\\
-------------------------------\\\\$

$\bf \cfrac{smaller}{larger}\qquad \cfrac{h}{h}=\cfrac{\sqrt[3]{volume}}{\sqrt[3]{volume}}\implies \cfrac{1}{2}=\sqrt[3]{\cfrac{volume}{volume}}
\\\\\\
\left( \cfrac{1}{2} \right)^3=\cfrac{volume}{volume}
\implies 
\cfrac{1^3}{2^3}=\cfrac{volume}{volume}\implies \cfrac{1}{8}=\cfrac{volume}{volume}\\\\
-------------------------------\\\\
\cfrac{smaller}{larger}\qquad \cfrac{1}{8}=\cfrac{\stackrel{volume}{v}}{\stackrel{volume}{400}}\implies \cfrac{1\cdot 400}{8}=v$

7 0

3 years ago

Identify the least common multiple that would eliminate the y variable in the system of equations 6x-5y=-4 4x+2y=28

nikdorinn [45]

The coefficient for y is -5 in the first equation, +2 in the second equation, so the least common multiple is -10

8 0

3 years ago

Question 1

fiasKO [112]

The scale factor is 3

4 0

3 years ago