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Nasty, Sneaky Santorum Ad Equates Obama With 'America's Enemies' (crooksandliars)

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This is Rick Santorum's clever new "Obamaville" ad. Look at the subliminal
editing of Barack Obama into a shot of Iranian President Mahmoud Ahmadinejad
while the voiceover talks about America's enemies. (h/t Digby.)

There's a good reason why we rejected Rick Santorum by 17 points in
Pennsylvania. Just like Doug Neidermeyer in "Animal House," he's a sneaky
little sh*t. He cooked up an elaborate charade that his family was still
living in his old Pennsylvania house for several years, merely so he could get
the local school district to cough up $72,000 for his kids' homeschooling
extras while they lived in their Leesburg, Virginia manse.

enlarge

There was also some very interesting monkey business in just how he got the
mortgage:

> The family returned to Fairfax County in August 2007. They bought a house
with five acres in Great Falls **with a high-ranking official of a major
development and mortgage company.**

>

> Santorum formed the Creamcup Trust with James Sack, the secretary and
general counsel of NVR, a major single-family developer and mortgage finance
company in northern Virginia and 15 states. Creamcup Trust bought a house and
five acres of land on ...

crooksandliars

can someone please explain to me how to do these math problems?

here's the link to the picture file...

http://s163.photobucket.com/albums/t286/spiffy-t-monkey/school%20stuff/?action=view&current=mathproblemhelp.jpg

then just click on the picture....

Hi,

For #1, There are 2 right angles at the bottom of "z", so both small triangles are right triangles. The entire large triangle all has a right angle at the top.
To start, the little triangle on the left is a 30-60-90 right triangle since the side of length 3 is half the hypotenuse of 6. That means the side of length 3 is across from the 30 degree angle. That means the angle in the bottom left corner is 60 degrees and the side across from it is equal to the l;ength of the 30 degree side times the square root of 3. That means that z = 3√3.
Since the angle in the bottom left corner is 60 degrees and
the entire large angle at the top is a right angle, that means the angle in the bottom right corner is 30 degrees. So in the 30-60-90 triangle on the right side, the 30 degree side is 3√3, the previously found "z". The hypotenuse is twice as long as this side, so "y" = 6√3. That means that "x" is the 60 degree side which equals the 30 degree side. 3√3 times √3, so x = 9.

In #2, the line in the interior of the triangle bisects the angle at the top. There is a theorem that says this angle bisector line divides the opposite side of the triangle into lengths proportional to their adjacent sides. So if the entire length on the bottom is 6 and the left end has length "x", we will call the right hand length "6 - x" instead of "y". Then we can write the proportion of the left lengths to the right lengths as:

..5......4
----.=.-----
.x......6-x

Cross-multiplying,

5(6-x) = 4x
30 - 5x = 4x
30 = 9x
10/3 = x

y = 6 - x = 6 - 10/3 = 8/3

For #3, let's call the right hand part of the base "w", so the left hand part would be the remaining 13 - w. Using these lengths, we could write 2 equations for the sides of both small right triangles using the Pythagorean Theorem:

5² = x² + (13 - w)²
12² = x² + w²

These simplify to:

25 = x² + 169 - 26w + w²
144 = x² + w²

Solving these both for x²,

x² = -144 + 26w - w²
144 - w² = x²

Since these expressions both equal x², they equal each other:

-144 + 26w - w² = 144 - w²

This simplifies to:

-144 + 26w = 144
26w = 288
w = 288/26
w = 144/13

The left hand side of the base is the remaining 25/13.

To solve for x, solve either equation. To be sure, solve both.

5² = x² + (13 - 144/13)²
25 = x² + (25/13)²
25 = x² + 625/169
25 = x² + 625/169
4225 = 169x² + 625
3600 = 169x²
3600/169 = x²
60/13 = x

12² = x² + w²
12² = x² + (144/13)²
144 = x² + 20736/169
24336 = 169x² + 20736
3600 = 169x²
3600/169 = x²
60/13 = x

It's so nice to get the same answer both ways!

On #4, Pythagorean Theorem time:

c² = a² + b²
......__
(3√10)² = 9² + x²
90 = 81 + x²
9 = x²
3 = x

On #5, the figure is a right triangle, but it also has congruent acute angles, so it's a 45-45-90 triangle. With 2 equal angles, there are also 2 congruent sides, so x and y will be equal. When you know the hypotenuse, you can find the length of a leg by dividing the hypotenuse by √2. In this case,

14........._
---- = 7√2
√2

For #6, the triangle is a 30-60-90 degree triangle where the sides have the lengths s-s√3-2s. Since the hypotenuse is 8, then 2s = 8, so s = 4. That's the value of x, the 30 degree side. That means that the 60 degree side is 4√3. That's the value of y.

4 = x
.....__
4√3. = y

For #7, the slope is 4, which is the tangent of the angle formed by the line and the x axis. To find the angle, find tan^-1(4) = 75.96 degrees

For #8, the slope is 1/2, which is the tangent of the angle formed by the line and the x axis. To find the angle, find tan^-1(1/2) = 26.57 degrees

Guessing for #9 and #10, you want the lengths in the x and y direction for these vectors.

#9. x = 128 cos(42) = 92.89
y = 128 sin(42) = 83.64

#10 x = 285 cos(65) = 120.45
y = 285 sin(65) = 258.3

#11 Assume that the quadrilateral at the top is a parallelogram. Then the horizontal line through the figure has to have a length of 15, just like the top. The triangle on the left and the triangle at the bottom are similar triangles, so their corresponding sides are proportional. This gives:

..9.......x
----.=.-----
10......15

Cross multiplying,

10x = 135
x = 13.5

#12 Since the bottom is 4 more than the top base of the trapezoid, then there is a 45-45-90 triangle in the right hand end, which means the height is also 4. The area of the trapezoid is found by:
A = .5h(b1 + b2)
A = .5(4)(3 + 7) = 2(10) = 20

#13 The formula for the area of an equilateral triangle is
A = s²√(3)/4. This becomes:
.............__...............__..........._
A = 6²√(3)/4 = 36√(3)/4 = 9√3 sq ft

#14 If the base is cut in half, then each side of the isosceles triangle forms a right triangle with a leg of 12 ft and a hypotenuse of 20 ft. Using the Pythagorean Theorem,

c² = a² + b²
20² = 12² + b²
400 = 144 + b²
256 = b²
16 = b

16 is the height and 24 is the base of the triangle, so

A = 1/2(24)(16)
A = 192 sq ft

#15 On each end of this isosceles trapezoid is a right triangle with hypotenuse of √29 and a leg of 2, half the additional length of the bottom base. Pythagorean Theorem, here we come!

c² = a² + b²
(√29)² = 2² + b²
29 = 4 + b²
25 = b²
5 = b This is the height of the trapezoid. So its area is:

A = .5h(b1 + b2)
A = .5(5)(8 + 12)
A = .5(5)(20)
A = 50

#16 A regular octagon can be divided into 8 isosceles triangles with an apthem or height of 9 feet. Since the 360 divided into 8 triangles makes each vertex angle 45 degrees, then the apothem divies that angle in half when forming 2 right triangles, with an angle of 22.5 degrees and an adjacent side of length 9 feet. Solve for the length of the opposite side by saying
.......................x
tan(22.5) = --------
.......................9

So 9 tan(22.5) = x and x = 3.728. That is half of the isosceles triangle's base, which is really 7.456. the area of each of the 8 triangles is A = .5(7.456)(9) = 33.55. The area of the octagon is 8 times as large or 286.4 sq. ft.

#17 A regular pentagon divides into 5 isosceles triangles with a vertex angle of 360/5 or 72 degrees. Dropping an apothem or altitude down through each triangle froms 2 right triangles with an acute angle of 72/2 or 36 degrees. Since the perimeter of the pentagon is 50 cm, then each side is 10 and a half-side is 5 cm. So in our 36 degree right triangle, we can find the height, the adjacent side, by saying:

...................5
tan 36 = ----------
....................h

Solving for h, h = 5/(tan 36) = 6.882

The area of each of the 5 triangles is A = .5(6.882)(10) = 34.41 sq cm. The area of the pentagon is 5 times as much or 172.1 sq cm

#18 The area of the entire square is 12² or 144 sq units. The area of the circle is Pi*r² or Pi*6² = 113.1 sq units. The shaded area is 144 - 113.1 = 30.9 sq units in the shaded area.

30.9 out of 144 sq units equals .2146 or 21.46% likelihood of being in the shaded area.

Wow, you win the longest problem award!!!

I hope this helps!!