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Jan 21, 2018
 #6
avatar+2345 
0

Here is an extremrly basic mockup of a number table from 0-99 for you to reference as I make the adding process easier.

 

 

I notice a few patterns here. The tens digit remains the same. Let's attempt to add the numbers 0-9.

 

\(\underbrace{0+\underbrace{1+\underbrace{2+\underbrace{3+\underbrace{4+5}+6}+7}+8}+9}\\ \)

 

The numbers enclosed by the underbraces sum to nine (9+0=9, 1+8=9, 2+7=9, 3+6=9. 4+5=9). There are 5 lots of this occurring with the numbers one to nine, so the sum of the tens digit in the first columns equals \(9*5\text{ or }45\). This phenomenon occurs ten times or once per column, so let's multiply this by 10. \(45*10=450\)

 

Now, let's do the ones column. The ones column is different because each column increments the amount by 1. The sum of the ones column can be represented by the following sequence. \(0*10+1*10+2*10+...+8*10+9*10\). We can use algebra to group the common factor, so it simplifies to \(10(1+2+3+...+8+9)\). Of course, we already know that the sum of the numbers 1-9 sums to 45 by our first calculation. 10 times that amount yields \(450\)

 

Of course, we have only dealt with the sum of the digits from 0-99. However, if you think about it, the sum of 100-109 should be the same as adding the ones digit and the hundreds digits. Of course, this is a special case since the tens digits are all 0. We already know that \(1+2+3+...+8+9=45\). We know that the hundreds digit of the numbers 100-109 sums to \(1*10=10\). However, we have forgotten 110, which has a value of 2. \(45+10+2=57\) for the sum of the digits from 100-110. Now, let's add everything together.

 

\(450+450+57=957\) or the sum of the digits from 0-110

Jan 21, 2018
 #1
avatar+106515 
+1

1) Find the area of a regular 12-gon inscribed in a unit circle.

 

We have 12 identical triangles  identical isosceles triangles with equal sides of 1 and whose apex angle  between these sides =  360/12  = 30”  

 

The  area will  be given  by    (1/2) (1)^2*sin (360/12°)  =

 

(1/2 sin (30°)  =  1/2  *  1/2  =   (1/4 ) units^2

 

 

2) A regular hexagon has side length 6. If the perimeter and area of the hexagon are p and A, respectively, what is the value of (p^4)/(a^2)?

 

The perimeter, p, is 36  ⇒   p^4  =   36^4

The area, a,  is   (1/2)6^2sin (60)  =  18*√3/2  =  9√2  ⇒ a^2  = (√182) ^2   = 162

 

So p^4 / a^2  =    36^4  / 162   =   10368 

 

 

3)Isosceles triangle OPQ has legs OP = OQ, base PQ = 2, and and angle POQ = 45 degrees. Find the distance from O to PQ.

 

 

The distance from O to PQ  is the altitude....call the point where the altitude intersects the base, R

And this altitude bisects POQ....so angle POR  =  22.5°

And the altitude also bisects the base.... so PR  =  1

 

Using the tangent function....we have that

 

tan (22.5)  =  1 / altitude        rearrange as

 

altitude  =   1  / tan (22.5)  ≈ 2.4142  =    1 + √2   

 

 

 

4) A, B, C, D and E are points on a circle of radius 2 in counterclockwise order. We know AB = BC = DE = 2 and CD = EA Find [ABCDE].
Enter your answer in the form x + y√z in simplest radical form.

 

 

This is a regular pentagon inscribed in a circle....I assume you want the area of ABCDE??

If so...... the area is   5 (1/2)(2)^2sin (72)  = 10sin (72)  = 10 √  [  5/8  + √5/8 ]  units^2

 

If you want the perimeter...we  have that  the half side lengh  = 

2sin(36)

And we have 10 half side lengths comprising the perimeter....so....the perimeter  =

10 * 2 *  sin(36)   = 20√ [ 5/8  - √5/8 ] units

 

 

cool cool cool

Jan 21, 2018
 #7
avatar+1439 
+1
 #2
avatar+106515 
0
Jan 21, 2018

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