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Definitely not that way..... Calculators are not allowed in the competition......

You will have to use the Law of Cosines...and then the Law of Sines...calling the sides A, B  and C  and the angle beween sides A  and  B   =  θ  =  angle C

C^2  = A^2  + B^2  - 2(A)(B)cos [θ]       rearrange

cos [θ]  =  [ A^2 - B^2 - C^2] / [ 2(A)(B)]   = θ   

Now......use the cosine inverse to find  θ

arcos (  [ A^2 - B^2 - C^2] / [ 2(A)(B)]  )   = θ  = angle C

Now.....use the Law of Sines to find another angle.....

sin [angle C] /C  =  sin B / B       rearrange

sin B  =  B* sin[angle C] / C

And B  can be found using  the  sine inverse

arcsin [ B* sin[angle C] / C ]  = angle B

And angle A  = 180 - B -  C

There are probably easier ways to do this...... but

We have three "outside" triangles and one inner one

Area  of the the three "outside" triangles  =

(1/2)(√3/2)  [ 6*4 + 2 * 4  + 6 * 2)  ]  =  11√3   units^2

And we an construct an inner triangle with sides BF, BD and FD

Length of BF  =  √[ 6^2 + 2^2 – 2(6)(2)(1/2)]  = √28  

Length of BD  =  √[ 6^2 + 4^2 – 2(6)(4)(1/2)]  = √28

Length of FD  =  √[ 2^2 + 4^2 – 2(2)(4)(1/2)]  = √12

Deriving the semi-perimeter of this triangle →     s  =  [2√28 + √12] /  2  =

√3 + 2√7  = √3 + √28     →  s^2  =  31 + 2√84  = 31 + √336

And using Heron's formula  ...

Area of inner triangle = √ [ (√3 + √28)(√3 + √28 – √28)^2 (√3 + √28 – √12)  ]    =

√ [ (√3 + √28)(√3 )^2 (√3 + √28 – √12)  ]  =

√3   *  √ [ (√3 + √28) ((√3 + √28  – √12)  ]  =

√3   *  √ [ (s) ((s  – √12)  ]  =

√3   *  √ [ s^2  – √12 s)  ]  =

√3   *  √ [ 31 + √336 – √12 [ √3 + √28)  ]  =

√3   *  √ [ 31 + √336 – √36 - √336)  ]  =

√3   *  √ [ 31  – 6   ]  =

√3   *  √ [ 25   ]  =

√3   *  5   =

5√3  units^2

So....the total  area  = [11√3  + 5√3]   units^2   = 16√3   units^2

"Benchmark", generally means "something that can be used as a way to judge the quality or level of other, similar things"!!! So, what do YOU mean by "benchmark"???!!.

PMT = PV x R {[ 1 + R]^N / [1 + R]^N - 1}

PMT =19,264 x 0.072/2{[1 + 0.072/2]^(6*2) / [1 + 0.072/2]^(6*2] - 1}

PMT = 693.504 x                            2.8914973......

PMT = $2,005.26

Solve for t over the real numbers:
297.49 e^(-0.0144009 t) - 140.83 = 0

297.49 e^(-0.0144009 t) - 140.83 = 29749/100 e^(-(25 t)/1736) - 14083/100:
29749/100 e^(-(25 t)/1736) - 14083/100 = 0

Bring 29749/100 e^(-(25 t)/1736) - 14083/100 together using the common denominator 100 e^(25 t/1736):
-1/100 e^(-25 t/1736) (14083 e^((25 t)/1736) - 29749) = 0

Multiply both sides by -100:
e^(-25 t/1736) (14083 e^((25 t)/1736) - 29749) = 0

Split into two equations:
e^(-25 t/1736) = 0 or 14083 e^((25 t)/1736) - 29749 = 0

e^(-25 t/1736) = 0 has no solution since for all z element R, e^z>0:
14083 e^((25 t)/1736) - 29749 = 0

Add 29749 to both sides:
14083 e^(25 t/1736) = 29749

Divide both sides by 14083:
e^(25 t/1736) = 29749/14083

Take the natural logarithm of both sides:
(25 t)/1736 = log(29749/14083)

Multiply both sides by 1736/25:
Answer: |t = 1736/25 log(29749/14083)= 51.92911533539

Volume=Pi x r^2 x h

Diameter/2 =Radius

r = [24.25 x 2.4] / 2 =29.1 ft - the radius

V =3.14 x 29.1^2 x 24.25

V = 64,480.35 Cubic feet - the volume of the cylinder.

P.S. If you use Pi to several decimal places, it gives you this volume =64,513.05 ft^3.

15. In ∆ ABC, D is a point on AB such that CD bisects ∠ACB, ∠A = 2∠B, AC = 11 and AD = 2

      Find the length of BC.

By Euclid......BC/DB  = 11/2  = 5.5   so  BC  = 5.5DB

And  < A  = 2 <B 

So

sinA/ BC  = sin B /11

sin (2B) /BC  = sin B / 11

[2sinBcosB] / BC  = sinB/ 11

2cosB/BC  = 1/11 

22cosB = BC

cosB  = BC/22 = 5.5DB/22   =  DB/4  

Using the Law  of Cosines....we have

11^2  = (BC)^2 + (DB+2)^2  – 2(BC)(DB+2)cosB

11^2  =  (5.5DB)^2 + (DB+2)^2 – 2(5.5DB)(DB+2)(DB/4)

let x  = DB

121  = 30.25x^2 + (x + 2)^2  – 2.75x^2 * (x + 2)

121 = 30.25x^2 + x^2 + 4x + 4 – 2.75x^3 – 5.5x^2

2.75x^3 - 25.75x^2  – 4x + 117  = 0

275x^3 – 2575x^2 –400x + 11700  =  0

Solving  this for x  produces two  possible positive  solutions   x = 26/11  or x = 9

So  DB  = 26/11   or DB  = 9     →   BC  = 5.5(DB)  = 5.5(26/11) = 13  

Or

BC  = 5.5(9)  = 49.5

But  BC  cannot  = 49.5  because cosB  = BC/22  = 49.5/22  which is impossible

So.......BC  = 13

if you buy 43 liters of gasoline for $40.64, how much for 1 liter.

Just divide: $40.64 / 43 =$0.945 - cents per Liter.

(3/4) / (5/12) =(3/4) x (12/5) =36 / 20 =1 4/5

1) - The common difference between the angles =44/(5-1) =11 degrees.
Sum =N/2[2F + (N - 1)*D]
540 =5/2[2F + (5 -1)*11], solve for F
F =86 degrees - the smallest angle.
86+44 =130 degrees - the largest angle.

if y=17 and x=45, using the formula y=kx, what does k=?

y = kx

17 =45k, divide both sides by 45

k =17/45 - and that is it.

19. Find the highest(greatest) common factor of 3^2046 - 1 and 3^2016 - 1.

Wolfram/Alpha gives =728. I have NO IDEA how they got it short of factoring each number individually, which too much to expect for such large numbers!!!???. Can anybody explain it? Thanks.

8.

There are 2016 different proper fractions. The sum of  any 2015 of them is a proper fraction with denominator 2017 when expressed in its simplest form. Find the sum of these 2016 proper fractions.

CPhill: Does this form an arithmetic series with the first term=1/2017..........and the last term=2016/2017, the common difference =1/2017, number of terms =2016???. If so, then it sums up to =1,008???.

\((-2 + \sqrt{-12}) + (8 +\sqrt{-27})\)

= \(-2+2\sqrt{-3} + 8 + 3\sqrt{-3}\)

Combine like terms.

=\(6+5\sqrt{-3}\)

The square root of negative one equals i, so take that out.

= \(6+5i\sqrt{3}\)

11)

M=1! x 2! x 3!........10! =6,658,606,584,104,736,522,240,000,000.

[6,658,606,584,104,736,522,240,000,000]^1/3

=1,881,313,269 Perfect cubes!!.

Cphill: Please check this! Is there a shortcut to this? Thanks.

What is -6 and 1/3 divided by -3.

[-6 1/3] / -3 =[- 19/3] x -1/3 =19/9 =2 1/9

Here are the last 10 or so digits........... 2899855929.

7.   The smallest  integer  Pythagorean triple is when AD  = 3 , AB  = 4 and DB  = 5

Perimeter of ABD  =  12

Area of ABCD  = AB * AD  =  [ 3 * 4 ]    = 12

Scaling ABCD  up by  a factor of 2  produces   AD  = 6, AB  = 8  and DB  = 10

Perimeter of ABD  =  24

Area  of ABCD  = [ 6 * 8 ]  = 48  = minimum value of S  for integer sides

tangent squared

Let  P/6 be the side length

Area  =  6 *(1/2)(P/6)^2*sin(60)  = 3(P^2/36)√3/ 2   = [ 3√3/72 [ P^2  = [ √3/24 ]P^2

So

P^2 /A  =   P^2 /  [ √3/24 * P^2]  =   1 / [√3/24]  =    24 / √3   =    [ √3 * 24] / [ √3 *√3] =

√3 * 24  / 3  =   8√3 

Part (a)

f(x) = (ax + b)/(x + c)

f(infinity) = - 4  so  - 4 = a       (limit as x goes to infinity of (ax+b)/(x+c) is a)   so a = -4

f(3) = infinite so  3 + c = 0     so c = -3

f(1) = 0        so  (a + b)/(1 + c) = 0   

                          (-4 + b)/(1 - 3) = 0      so b = 4

hence f(x) = (-4x + 4)/(x - 3)      or    f(x) = 4(1 - x)/(x - 3)

I'll leave you to do part (b).

A regular hexagon is comprised of 6 equilateral triangles. Each side is s = p/6

Area of 1 triangle:  A1 = (1/2)*s*sqrt(s^2 - (s/2)^2) → (1/2)*s^2*sqrt(3/4) → (1/72)*p^2*sqrt(3/4) 

Area of hexagon;  A = 6A1 → (6/72)*p^2*sqrt(3/4) → (1/12)*p^2*sqrt(3/4) → (1/24)*p^2*sqrt(3)

Hence p^2/A = 24/sqrt(3) → 8sqrt(3)

.

Thanks Alan, this confirmed the answer I got two different ways (interpolation between 8.20 and 8.30 and 8+25.5/60 (virtually the same as your way) = .14651466