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2 years ago

3. WEATHER On a cold January day,

Mathematics

1 answer:

erma4kov [3.2K]2 years ago

3 0

Answer: -9 + 21 = t t = 12°C.

Step-by-step explanation: For this problem lets have t be equal to the original temperature in the beginning of the day. The equation -9 + 21 = t represents this situation because it states that if we add the degrees that dropped to the current degrees it will give us our original degrees.

Solve.

-9 + 21 = t (Simply add -9 + 21) -9 + 21 = (+12)

12 = t

So, the original degrees in the beginning of the day was 12°C.

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3 years ago

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2 years ago

lbvjy [14]

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

$\bf \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{48\pi }}{\sqrt{75\pi }}\implies \cfrac{s}{s}=\cfrac{\sqrt{(2^2)^2\cdot 3}}{\sqrt{5^2\cdot 3}}\implies \cfrac{s}{s}=\cfrac{4\sqrt{3}}{5\sqrt{3}}
\\\\\\
\cfrac{s}{s}=\cfrac{4}{5}$

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