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3 years ago

if a motercycle is moving at a constant speed down the highway of 40 km/hr, how long would it the motorcycle to travel 10 km

Mathematics

1 answer:

ahrayia [7]3 years ago

6 0

Answer:

15 minutes

Step-by-step explanation:

First, the motorcycle goes at a speed of 40 km/hr.

The question asks for how long it would take to travel 10 km.

Well, there are 60 minutes in an hour, since we will be translating into minutes.

Also, 10 km is 1/4 of 40 km, so it would make sense that the time length would be 1/4 of an hour as well.

1/4 of 60 minutes is 15 minutes. So it takes 15 minutes for the motorcycle to travel 10 km.

Now, if all this wordy stuff is too much to comprehend, you can also solve using proportional relationships.

$\frac{40km}{60min}=\frac{10km}{xmin}$

Now cross multiply:

$40km*xmin=10km*60min\\40x=600$

Divide both sides by 40:

$\frac{40x}{40}=\frac{600}{40}\\x=15$

Again, this shows that it wouls take 15 minutes for the motorcycle to travel 10 km.

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


\bullet \textit{ vertical shift by }{{ D}}\\
\qquad if\ {{ D}}\textit{ is negative, downwards}\\\\
\qquad if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}
\end{array}$

now, with that template above in mind, let's see this one

$\bf parent\implies f(x)=|x|
\\\\\\
\begin{array}{lllcclll}
f(x)=&3|&1x&+2|&+4\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}$

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check the picture below

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