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Math Homework Help

Geometry in Court

published on December 4th, 2006 . by Vanaja

In real life, math can be relevant in the courtroom or come up in legal disputes.

Last year, the Pythagorean theorem was a deciding factor in a case before the New York State Court of Appeals. A man named James Robbins was convicted of selling drugs within 1,000 feet of a school. In the appeal, his lawyers argued that Robbins wasn’t actually within the required distance when caught and so should not get the stiffer penalty that school proximity calls for.

The arrest occurred on the corner of Eighth Avenue and 40th Street in Manhattan. The nearest school, Holy Cross, is on 43rd Street between Eighth and Ninth Avenues.

Law enforcement officials applied the Pythagorean theorem to calculate the straight-line distance between the two points. They measured the distance up Eighth Avenue (764 feet) and the distance to the church along 43rd Street (490 feet), using the data to find the length of the hypotenuse, 907.63 feet.

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UW wants to boost math enrollment

published on December 4th, 2006 . by Vanaja

On the surface, math may seem less glamorous than biology or engineering, but the field, which deals with abstract concepts and equations, is integral to all other scientific fields.

As the United States faces a critical shortage of American scientists, the University of Wisconsin-Madison will lead an effort to attract more students to mathematics, ultimately providing the sciences with a stronger, smarter workforce.

How Can We Draw The Graphs Of Trigonometric Functions?

published on December 3rd, 2006 . by Vanaja

We have four ways to alter the graph of a function: vertical translation, vertical scaling, horizontal translation, and horizontal scaling. Note that some or all four of these can be applied to a function at once in the sense that you may have to deal with a function of the form y = AsinB(x - C) + D. You should simply work these graphs out one step at a time, concerning yourself first with the affect of B and C, then the affect of the presence of A, and finally the affect of D.

Example:
Draw the graph of y =2 sin(x/2+?/6)+3

Solution
Let’s change this is of the form y = AsinB(x - C) + D
So y =2 sin(x/2+?/6)+3
can be written as y =2 sin ½ [x-(-?/3)]+3
Step 1: Draw sinx


B= ½ .so period is 4pi. That means it completes one cycle after 4pi.so stretch the graph horizontally so that its period is 4pi.

Next, we see that C=-pi/3. This will horizontally shift our graph pi/3units to the left. Thus, we now have


Now A=2, which will vertically scale our graph by a factor of 2. Thus, at this point, our range should now be [-2,2]

Finally, we must shift our graph vertically 3 units due to the presence of D=3. Hence, our range will now move to [1,5]

We can adopt the same method for drawing the graphs of other trigonometric functions. But remember first you should convert the given functions of the form

y = A sinB(x - C) + D

y = A cosB(x - C) + D

y = A tanB(x - C) + D

or accordingly, depending upon which of the trigonometric function is given.

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