What is the equation for the relationship

Mathematics

1 answer:

Nostrana [21]3 years ago

6 0

The equation is y=0.4 i think

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$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array}\\\\
-----------------------------$

$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\
\stackrel{\stackrel{pyramids}{scale~factor}}{\cfrac{small}{large}}\qquad \qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{64}}{\sqrt[3]{343}}\implies \cfrac{s}{s}=\cfrac{4}{7}\implies 4:7$

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If 4c-2=c then 12c= what ??

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Answer:

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Step-by-step explanation:

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Water is being pumped into a 12-foot-tall cylindrical tank at a constant rate.

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Answer:

4 feet

Step-by-step explanation:

It is given that:

• At 2:30 p.m., the water depth was 2.6 feet.

• It is now 5:00 p.m., and the depth of the water is 3.6 feet.

So in 2.5 hours ( 5pm - 2:30pm) the water rose 1 feet (3.6 - 2.6).

Now <em><u>we can find how much water rises in 1 hour by setting up a ratio (let x be the depth increase of water in 1 hr):</u></em>

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So, in 1 hour, the water level will rise 0.4 feet

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