CPhill

a)  Here, we are trying to find angle PQR.....we can use the Law of Cosines for this...so we have...

3^2  =  6^2 + 4^2  - 2(6)(4)cosQPR    rearrange as

[ 3^2 - 6^2 - 4^2 ] / [ - 2(6)(4)]  =  cos QPR

Using the cosine inverse (arccos), we have that

arccos ( [ 3^2 - 6^2 - 4^2 ] / [ - 2(6)(4)] )  = QPR ≈  26.38°

So....the bearing of R from Q  =  [180 - 26.38]°  ≈  153.62°  = 154°

b) This one is easier...here we are trying to find angle QPR.....we can still use the Law of Cosines

4^2 =  6^2  +  3^2  - 2(6)(3)cos QPR   rearrange

([ 4^2 - 6^2 - 3^2 ] / [ -2(6)(3)] )  =  cosQPR

Using the cosine inverse (arccos), we have the bearing of R from P  is

arccos ([ 4^2 - 6^2 - 3^2 ] / [ -2(6)(3)] )  = QPR  ≈  36.34° = 36°

Here's a pic :

We can use the Law of Cosines to solve this

We have that

96.8^2  = 130^2  + 56^2  - 2(130)(56)cos(ABC)       rearrange as

[ 96.8^2 - 130^2 - 56^2 ] / [ -2(130)(56)]  =  cos(ABC)

Using the cosine inverse (arccos), we have

arccos ( [ 96.8^2 - 130^2 - 56^2 ] / [ -2(130)(56)] )  = ABC ≈  42.9°

2)

Here's a pic of the way I set this up :

The radius  of the larger  hexagon can be found as follows :

Area   =  1            ....so....

3 r^2 sin (60)  =  1

3r^2 * √3/2 = 1

r^2  =    2  / √27

r  =  √  [ 2 / √27 ]

The y coordinate of  B  is      r * sin (60°) =  √  [ 2 / √27 ] * [ √3 / 2 ]  

And the y coordinate  of F  is  r * sin (-60°)   - √  [ 2 / √27 ] * [ √3 / 2 ]

And the distance between them, BF,  is just    2 * √  [ 2 / √27 ] * [ √3 / 2 ]    =  

√  [ 2 / √27 ] *  √3 =

√ [ 6 / √27]

Note that because they span equal arcs, chords AC and BF are equal

And by SSS, triangle CBA is congruent to triangle BAF  and both are isosceles

And angle ACB  = angle BAC =  angle BAF = angle FBA

So  angle FBA  =  angle BAC....so BH  = HA

And angle CBA  = angle BAE  =  90°

So...subtracting equal angles , then angle HBI  = angle HAM

And angle BHI  = angle AHM

So...by  SAS, triangle BHI  is congruent to triangle AHM

But  since HI  is a transversal that cuts parallels  DB  and AE....then angle HAM  = angle HIB

But angle HAM  =  angle HBI......so  angle HBI  = angle HIB

And angle HIB  =  angle HMA.....so angle HAM  = angle HMA

So  HM   = AH      and  HI  = HB

So.....HM = HI = AH = HB

So .....since triangles BHI and AHM are congruent.....then HM  = HB

And by similar logic, we can prove that HM  = MF

So..... HB  = HM = HF

So....side HM of hexagon HIJKLM  is 1/3  of BF

So....the area of the hexagon is

3 * [ 6 / √27] / 9 * √3 / 2  =

√3 / √27  =

√3 /  [ 3 √3]  =

1/3 units^2

3 and 4 answered here :

https://web2.0calc.com/questions/math-help-please_16

5)

\($\dfrac {1}{1\cdot 3} + \dfrac {1}{3\cdot 5} + \dfrac {1}{5\cdot 7} + \dfrac {1}{7\cdot 9} + \cdots + \dfrac {1}{199\cdot 201}$\)

Note that  we can write      1 /  [  n (n + 2) ]  as     (1/2)  [ 1/n  -  1 / (n + 2)]

So  we can write

(1/2) [  (1/1  - 1/3)  + ( 1/3 - 1/5) + ( 1/5 - 1/7) + ( 1/7 - 1/9)  + .....+ (1/197 - 1/199) + (1/199 - 1/201) ]  =

(1/2) (1/1  -  1/201)  =

(1/2) ( 201 - 1 ) / 201  =

(1/2) (200) / 201  =

100 / 201

1)

(1 - 1/2) * (1 - 1/3) * (1- 1/4) * (1 - 1/5) *...........* ( 1 - 1/2007) * (1 - 1/2008) * (1 -1/2009)  = 

(1/2) (2/3) (3/4) ( 4/5)  *...........*   ( 2006/2007) (2007/2008) (2008/ 2009) =

(1) (2/2) (3/3) (4/ 4) (5/5) *.........* ( 2006/2006) (2007/ 2007) (008/ 2008) (1/2009) =

(1) (1) (1) *...... *(1/2009)  =

(1 / 2009)

Here's the first one:

See the way I've set it up

angle ABC  =  120°  ......so the area of triangle ABC  =  (1/2)*2^2 sin (120) =  √3 units^2

And ACDE forms an isosceles trapezoid

Let the bottom base be  DE  = 2

A  = (0,2)    and C  =  (-√ (3) , -1)    E  = (2, 0)    D =  ( 1, -√3)

And  AC forms the top base  = √  [   3 + 9  }  =  √12 units

Call G the mid-point of  AC  =  (-√3/2 , 1/2 )

And  F  is the midpoint  of  DE  =  ( 3/2, -√3/2 )

And FG  is the height of the trapezoid  =  √ [  ( 3 + √3)^2  + (1 + √3)^2 ]  /  2  = 

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

So.....the area of the trapezoid  is

(1/2) (height)  (sum of the bases)  =

(1/2) ( 1 + √3)  [ 2  + √12 ]  =

(1/2) (1 + √3) [ 2 + 2√3 ]  =

( 1 + √3)^2  =   [ 4 + 2√3 ] units^2

So.....the total area is  

[ √3  + 4 + 2√3 ]    =    4 + 3√3  units^2

If tea has twice as much caffeine as coffee per volume,  then 160 cups of tea must have twice the amount of caffeine as 40 cups of coffee. So.....80 cups of tea must have the same amount of caffeine as 40 cups of coffee. So....one cup of coffee must have twice the caffeine of one cup of tea.

One cup of tea has 1/2 the caffeine as a cup of coffee. So.......since a 12-ounce can of cola has 1/4 the caffeine of a cup of coffee, a cup of tea must have twice the caffeine of a 12-ounce cola.

Consecutive angles in a parallelogram are supplemental....so....

B + C  = 180

110  +  C   = 180      subtract 110 from both sides

C  = 70°

y  = k /x^2

We need to find k

And we know that

9 =  k / 1      multiply both sides by 1

9  = k

So......when y  =  16, we have

16  = 9 / x^2     rearrange as

x^2  =  9 /16        take the positive root

x  =  3 / 4