Given right triangle mno, which represents the value of cos(m)?

answerthe vertex form

step-by-step explanation:

a significant challenge in working with any kind of data is how to present this data to answer questions of interest to human observers. this challenge is especially critical for big data streams because of the volume, variety, or velocity of the data involved. data of large volume has the potential to overwhelm an observers ability to extract meaning from the data. how should the voting data for every us citizen be portrayed to be of interest to political scientists? how should all of the genome data for a species be represented to answer interesting biological questions? how can all of the works of romantic poets be presented for literary analysis? data of highly diverse variety is a challenge because the data forms of data might best be conveyed by different means. integrating different representations in a meaningful way is difficult. for example health-related information contains family history, genomic data, medical test results, descriptions of exercise patterns, records of diet and nutrition, list of stress factors. how can all of this health-related information even for a single individual be displayed for best use by doctors? data with high velocity is changing so rapidly that interesting or significant aspects of the data might be lost. for example, how can stock market data be depicted so that meaningful trends can be seen by a stock analyst? how can data on the movement of vehicles in an urban road system be exhibited so that traffic engineers can determine better ways to avoid traffic jams?

data visualization is often a useful way to present complex data. the adage that “a picture is worth a thousand words” reflects the power of the human visual system. presenting data in a visual form leverages the ability of our visual systems to us see patterns or to detect special cases of interest that are distinctive in some way. finding a good visualization is, of course, not always easy. the picture game “where’s waldo” shows how easy it is to create a visual representation that hides the feature of interest.

there are a rich variety of ways to visualize data in both two and three dimensions. for example, here is a gallerylinks to an external site. of visualization that can be produced by the tools we will be using. developing the knowledge and skills to build intricate and compelling visualizations is a study on its own. however, some basic forms of visualization are very useful for a number of “first step” explorations of big data. these basic forms are:

line graphs: showing how data changes over time (e.g. how the value of a stock changes over the trading days).

histograms: showing the distribution over a range (e.g., how many earthquakes are there in a month in each of the six units on the richter scale).

scatter plots: showing the relationship between two different factors (e.g., how the temperature and humidity for a given city are related),

bar charts: showing the comparative values in certain categories (e.g., display the average forecast temperature for new york, ny, blacksburg, va, and seattle,  

map plots: showing the distribution over geographic space (e.g., where on the globe have earthquakes in the last month occurred).

you can also think of visualizations that you see every day as other examples of how data is presented: weather maps, hurricane tracks, election maps.

we will be using matplotlib to illustrate visualization of data. matplotlib is widely used and offers the capability to generate many different kinds of visualization in python. we saw examples of its variety of visualization in the matplotlib gallery above. however, this utility comes at a price. matplotlib is complicated. fortunately, matplotlib is designed so that many common cases can be done with minimal or reasonable effort. in addition to matplotlib we will also look at a library for geographical plotting that is built on top of matplotlib.

step-by-step explanation:

here we can see that the storage increases

at certain times of the year - during the

warmer months, when the gas is not much

used for heating and reserves can be

increased - and then decreases during the

colder months because the supply is being

used. it is possible to model this with a

relatively simple trigonometric graph, but

first we examine the simplest trig curves.

065/065.html

the basics

the simplest sine and cosine curves can be generated by plotting the trig-function values

against the angles used in each function. the values can be found in the chart below:

ø (˚) 0 30 45 60 90 120 135 150 180 210 225 240 270 300 315 330 360

sin ø

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0 1

2

2

2 3

2 1 3

2 2

2 1

2

0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0

cos ø

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1 3

2 2

2 1

2 0 − 1

2

− 2

2 − 3

2 − 1 − 3

2 − 2

2 − 1

2

0 1

2 2

2 3

2 1

first we plot the sine data, letting the x-axis represent the angles and the y-axis the sine

values. the result is the scatter plot below.

based on just these 17 data points, we can see

that the function is following a straightforward

pattern - the sine values are increasing to 1,

decreasing through the same values to -1, then

returning to zero. when we connect the dots, we

get the periodic curve y = sin (x) below.

this “sine curve,” with equation

y = sin(x), is the simplest of trig

graphs. the curve rises and falls by

one unit from its “axis of

oscillation” - in this case, the x-axis

- and after we cycle through 360˚ it

appears we are ready to go through

the same y-values again. we can see

this same pattern in the cosine

function on the following page.

each sine and cosine curve has two basic characteristics:

amplitude - distance the curve rises above, and falls below, its axis of oscillation (marked by

the vertical arrows on the cosine graph). the formula is

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amp = max− min

2 .

period - the length of one cycle (marked by the horizontal arrows on the cosine graph).

this can be calculated by measuring the distance between any successive maximum

points (or minimum points), or by doubling the horizontal distance between a

minimum and the next maximum.

for our first two curves, then, the amplitude is 1 and the period is 360˚ (or 2! if we are in radians).

adjusting the amplitude and frequency: y = a sin(bx), y = a cos(bx)

now that we have our “base” amplitude and frequency . . let’s change them. we can adjust both

characteristics by introducing the parameters a and b into the equations. the graph below shows the

curves y = a sin(x) for a = 1,2,3, and -0.5, with the original curve y = sin(x) being the thickest line.

clearly the amplitude of the curves has changed;

when a = 2, the amplitude is 2; when a = 3, the

amplitude is 3; and when a = -0.5 the amplitude is

0.5 (since amplitude, a distance, must be positive).

however, this last curve has also been flipped upside

down (or “reflected in the x-axis”). since the

amplitude must be positive even when a itself is

negative, we have the formula

amplitude = | a |.

the relationship between the b-value and the

period is not so straightforward. on the right

we have y = cos(bx) for b = 1,3, and 0.5. (the

thickest line is again our “base” curve.) when

b=3, the period decreases to 120˚; when

b=0.5, the period increases to 720˚ (so only

half the curve is shown here). we can describe

this reciprocal relationship with the formula

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b = 360

period or

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b = 2π

period ,

using the latter if we are working in radians.

shifting horizontally and vertically: y = sin(x-h) + k, y = cos(x-h) + k

to shift (or “translate”) a curve along either the x- or y-axis (or both), we introduce the parameters

h and k. below we have the graph of y = sin (x - 30) + 2 and y = sin(x). (note that we have returned

to an amplitude of 1 and period of 360˚). our original

curve y = sin(x) had its peak at the point (90,1); the

shifted curve has its corresponding peak at (120, 3). in

the same way, the original curve had a minimum value

at (270, -1); now the minimum is at (300,1). and

finally, the axis of oscillation has shifted vertically

from the line x = 0 to the line x =2. in other words, our

new curve has been created by shifting y = sin(x) right

30 and u p 2. and, in general, we can create the

following shifts:

horizontal shift = h (left if h < 0, right if h > 0)

vertical shift/new axis of oscillation = k.

modelling data: y = a sin[b(x-h)] + k, y = a cos[b(x-h)] + k

we have seen how the basic sine and cosine curves can be adjusted: by changing the amplitude

and period, and by shifting the curves horizontally and vertically. a summary is below.

amplitude = | a | vertical shift/axis of oscillation = k

period =

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360

b

or

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2π

b