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Jan 21, 2018
 #6
avatar+1604 
+1

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

 #1
avatar+81061 
+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

CPhill 7 hours ago
 #2
avatar+81061 
0
CPhill 10 hours ago
 #1
avatar+10858 
+1

It will have a vertical axis and the equation looks something like:

 

(x-1)^2/3^2 + (y-2)^2/4^2= 1      I'll have to do a little more review and get back to you....

 

The CENTER of the ellipse is directly between the foci

so center =   1,2    (h,k)

 

Equation for an ellipse looks like:

 

(x-h)^2 / b^2   +  (y-k)^2 / a^2 = 1    where h.k is the center (much like the equation for a circle)

so we have:

 

(x-1)^2 /a^2   + (y-2)^2 /b^2 = 1   we can find b^2 by substituting the point given (4,2)

results in b^2 = 9    or b =3

 

(x-1)^2/3^2 + (y-2)^2/a^2  = 1      Now we need to find a^2

Now....from Purplemath (where I have been reading about ellipses) is this:

The three letters are related by the equation b2 = a2 – c2 or, alternatively (depending on your book or instructor), by the equation b2 + c2 = a2.

(Proving the relationship requires pages and pages of algebraic computations, so just trust me that the equation is true. It can also be shown — painfully — that b is also the length of the semi-minor axis, so the distance across the ellipse in the "shorter" direction is 2b. Yes, the Pythagorean Theorem is involved in proving this stuff. Yes, these are the same letters used in the Pythagorean Theorem. No, this is not the same as the Pythagorean Theorem. Yes, this is very confusing. Accept it, make sure to memorize the relationship before the next test, and move on.)

 

OK   then      b^2 +c^2 = a^2       we know b = 3    c=3 (didstance from center to one of the foci) so we have:

 

3^2 +3^2 = a^2       thus a^2 = 18    and  FINALLY, we have the quation of the ellipse in question:

 

(x-1)^2 / 3^2   +  (y-2)^2 / 18  =1              

 

(there PROBABY is an easier way !!!!   This was my thought process as I was reading through Purplemath learning about ellipses....you might want to read through it too ! )

Here is a desmos graph:

ElectricPavlov Jan 21, 2018 4:01:41 PM
 #1
avatar+81061 
+2

 

Probably a way easier way to do this with Geometry...but....I didn't see it...so.....

 

By SAS,   triangles BCF  and ABE  are congruent

 

Therefore angle EAB  = angle FBC

And angle DAE   = angle BEA

But angle DAE + angle EAB  = 90

But angle FBC +  angle FBA  = 90

So  angle BEA =  angle FBA

So...by  AA congruence....triangles BEG   and ABG are similar

But angle BGA  = angle EGB...so....each must = 90

 

So triangles BEG an ABG are right triangles

And AB  = 2BE

So  AG  = 2 BG

So AB^2  =  AG^2  + BG^2

AB^2  =  (2BG)^2  + BG^2

AB^2  =  5BG^2

So AB   = sqrt(5)BG    =  BC

And  (1/2)AB   =  [sqrt(5)/2] BG

cos angle ABG  = BG/ AB  =  BG / sqrt(5)BG   =  1/ sqrt(5)

sin angle ABG =2BG/sqrt (5)BG  =  2/sqrt(5)

And angle ABG  = angle CFB

So sin angle ABG  = sin angle CFB

But  angle CFB  and angle GFD are supplementary so their sines are equal

So....sin ABG  = sin GFD     

And GFD is obtuse...so   

So... cos GFD  =  -sqrt [  1  - sin^2(ABG) ]  =  -sqrt [ 1 -  sin^2(GFD) ]  = 

 - sqrt [ 1 - (2/sqrt(5))^2 ]  = -sqrt [ 1 - 4/5]    = -sqrt (1 /5)  =  -1/sqrt (5)

 

And

Triangles ABE  and BCF  are congruent by SAS

EA  = FB     BC  = AB     and BE  = CF =  (1/2)AB  = FD

EA^2  =  BC^2 + BE^2    .... so.....

FB^2  =  AB^2  +  (AB/2)^2

FB^2  =  5BG^2 + AB^2/4

FB^2  =  5BG^2 +  (5/4)BG^2

FB   =   BGsqrt  (5 + 5/4)

FB =  BGsqrt [ 25/4]  =  (5/2)BG

 

And  

FG  =  FB - BG  =     (5/2)BG - BG  =  (3/2)BG

 

So  using the Law of Cosines

 

GD^2  =  FG^2  + FD^2 - 2(FG)(FD)cosGFD

GD^2  =  FG^2  + (AB/2)^2 - 2(FG)(AB/2)cosGFD

GD^2  = (9/4)BG^2 +   (5/4)BG^2 - 2 (3/2)BG * ( sqrt(5)BG/2) * [ - (1/sqrt(5) ) ]

 

GD^2  =  (14/4)BG^2 + (3/2)BG^2

GD^2  =  (7/2)BG^2 +  (3/2) BG^2

GD^2  =  (10/2)BG^2

GD^2 =  5BG^2

GD  =  sqrt(5)BG

DG  =  sqrt(5)BG  =  AB

 

 

cool cool cool

CPhill Jan 21, 2018 7:11:36 AM

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