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1 year ago

Use the scale of 1cm=45 km

Mathematics

1 answer:

alex41 [277]1 year ago

3 0

Answer:

d = 7.4 cm is the distance between the two cities on the map.

Step-by-step explanation:

Let d = the distance measured on the map in centimeters (cm) between any two locations and let D be the actual distance in kilometers (km) between these same two locations.

Since the scale on the map is 1 cm = 45 km, then the ratio of the distance on the map to the actual distance is equal to 1cm/45 km; therefore the quotient of any measured distance d on the map divided by the corresponding much longer actual distance D would have to maintain (equal) this same ratio; In other words, we can set up the following proportion to solve for d given the actual distance D = 333 km as follows:

1 cm/45 km = d/D

Substituting, we get:

1 cm/45 km = d/333 km

d/333 km = 1 cm/45 km

(d/333 km)(333 km) = (1 cm/45 km)(333 km)

d = (333 km/45 km)(1 cm)

d = 7.4 cm is the distance between the two cities on the map.

Check:

1 cm/45 cm = d/333 km

1 cm/45 cm = 7.4 cm/333 km

1 cm/45 cm = 1 cm/45 km (Therefore, the distance ratio of the map has been maintained

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$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

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

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\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

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