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julia-pushkina [17]
3 years ago
13

A map is made to a scale of 1/4 inch is to 12 inches. What is the actual length of a room in feet whose length on the map is 3 3

/8 inches?
Mathematics
1 answer:
egoroff_w [7]3 years ago
5 0
In this example, we are to find the actual length of a room based from its length on the map by using ratio and proportion.
For our solution to be more clearer, let us turn the fractions into decimals.
1/4 inch = 0.25 in.
3 3/8 inches = 3.375 in.

So, for every 0.25 inches from the map there will be 12 inches in the room, and what we will find here is that how many inches are there in the room if there are 3.375 inches in the map.

Let x, be the length in the room,
\frac{0.25}{12} = \frac{3.375}{x}
Cross multiply the given fractions and you would get:
0.25x=40.5
x=162

So the length of the room, if the value of it in the map is 3 3/8 inch, is 162 inches.
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Base: z(x)=cosx Period:180 Maximum:5 Minimum: -4 What are the transformation? Domain and Range? Graph?
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Answer:

The transformations needed to obtain the new function are horizontal scaling, vertical scaling and vertical translation. The resultant function is z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right).

The domain of the function is all real numbers and its range is between -4 and 5.

The graph is enclosed below as attachment.

Step-by-step explanation:

Let be z (x) = \cos x the base formula, where x is measured in sexagesimal degrees. This expression must be transformed by using the following data:

T = 180^{\circ} (Period)

z_{min} = -4 (Minimum)

z_{max} = 5 (Maximum)

The cosine function is a periodic bounded function that lies between -1 and 1, that is, twice the unit amplitude, and periodicity of 2\pi radians. In addition, the following considerations must be taken into account for transformations:

1) x must be replaced by \frac{2\pi\cdot x}{180^{\circ}}. (Horizontal scaling)

2) The cosine function must be multiplied by a new amplitude (Vertical scaling), which is:

\Delta z = \frac{z_{max}-z_{min}}{2}

\Delta z = \frac{5+4}{2}

\Delta z = \frac{9}{2}

3) Midpoint value must be changed from zero to the midpoint between new minimum and maximum. (Vertical translation)

z_{m} = \frac{z_{min}+z_{max}}{2}

z_{m} = \frac{1}{2}

The new function is:

z'(x) = z_{m} + \Delta z\cdot \cos \left(\frac{2\pi\cdot x}{T} \right)

Given that z_{m} = \frac{1}{2}, \Delta z = \frac{9}{2} and T = 180^{\circ}, the outcome is:

z'(x) = \frac{1}{2}  + \frac{9}{2} \cdot \cos \left(\frac{\pi\cdot x}{90^{\circ}} \right)

The domain of the function is all real numbers and its range is between -4 and 5. The graph is enclosed below as attachment.

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