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Sin (2A)  =  5/13  ...... so  ....Cos(2A)  =  12/13

Using a half-angle identity

cos (A)  =  √  [  ( 1 + cos2A)  / 2 ]  =   √  [ ( 1  + 12/13) / 2 ]  =

√  [  (13 + 12) / 26 }     =   √  [ 25 / 26 ] =     5 / √26  =

(5/26)√26

 And we take the positive root since the cosine is positive on the requested interval

no problem any time

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sin 30 = h/12     h= 6

6^2 + (1/2b)^2) = 12^2

(1/2 b)^2 = 144-36 = 108

b=20.78

Area = 1/2 bh = 1/2  20.78  x 6 = 62.35 units^2

Area=1/2absinC

Area=1/2*12*12*sin120

Area=1/2*144*sin120

Area=72*sin120

Area=62.4cm²(1dp)

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.98058      You DO know there is a calculator on this website......Yes?

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This was an easy to type qeustion

SO

Next time could you please just type it in!

Find polynomial f(n) such that for all integers \( n \ge 1\) we have
\(3\left( 1\cdot2 + 2\cdot3 + \ldots + n(n+1) \right) = f(n).\)

\(\begin{array}{|rcll|} \hline f(n) &=& 3\Big( 1\cdot2 + 2\cdot3 + 3\cdot4 + \ldots + n(n+1) \Big) \quad & | \quad n(n+1) = n^2+n \\ &=& 3 ( 1^2+1+2^2+2+3^2+3 + \ldots + n^2+n ) \\ &=& 3 ( \underbrace{1+2+3+ \ldots + n}_{=\dfrac{(n+1)n}{2}} + \underbrace{1^2 +2^2 +3^2 + \ldots + n^2}_{=\dfrac{(n+1)n(2n+1)}{6}} ) \\\\ &=& 3 \left( \dfrac{(n+1)n}{2} + \dfrac{(n+1)n(2n+1)}{6} \right) \\\\ &=& 3 \left[ \dfrac{(n+1)n}{2}\left(1+ \dfrac{2n+1}{3}\right) \right] \\\\ &=& \dfrac{(n+1)n}{2}\left(3+ 2n+1\right) \\\\ &=& \dfrac{(n+1)n}{2}\left(2n+4\right) \\\\ &=& (n+1)n(n+2) \\\\ &=& n(n+1)(n+2) \\\\ &\mathbf{=}& \mathbf{ n^3+3n^2+2n } \\\\ \hline \end{array}\)

a.  x^2  -  2x  = 24      subtract 24 from both sides

x^2  - 2x  -  24   =  0       factor

(x - 6) (x + 4)  = 0

Setting each factor  to 0  and solving for x  we get that   x  = 6   or x  =  -4

b.  Note that the area of the triangle on the right  =  (1/2)x ( x - 2)

And this is the same area as the one on the left   = (1/2)x (x -2)

Adding these we get

(x - 2)  [  (1/2)x  + (1/2)x]   =  (x - 2) [ x  ] =  x^2  - 2x

So......this means that  (a)   is representing the area of the triangle as 24 units^2

Proof..... using the positive solution  found in "a" we have that  

(6)^2 -2(6)   =   36  -  12   =   24 units^2

c.   If the surface area  is 120 cm^2

We can  solve this

x^2  - 2x  =  120       subtract 120 from both sides

x^2  - 2x  - 120   = 0      factor

(x - 12) ( x + 10)  =  0

Setting each factor to 0 and solving for x  we get that  x  =  12  or x  = -10

Taking  the positive solution   AB  =  2x  =  2(12)   =  24 cm 

[0,5]

What is (f+g)(x)?

f(x) = 8x^2+16x+6

g(x)=x^3−3x^2−9

f +  g  =       8x^2 + 16x  + 6   + x^3 - 3x^2 - 9   =   x^3 + 5x^2  + 16x - 3 

What is (f⋅g)(x)

f(x)=x^4−9

g(x)=x^3+9 

f * g   =

 (x^4 - 9( (x^3 + 9)     use the distributive property

x^4 (x^3) + x^4(9)  + (-9)(x^3)  + (-9)(9)

x^7  + 9x^4   -  9x^3  - 81

The amount of the commission earned decreases between the fifth and eleventh week of employment.

A commission of $320 for a week was earned three times over the first 10 weeks.

The commission earned increased in the beginning of employment and after week 11.

relative minimum = -36,

relative maximum = 64

I worked the first one earlier.....I thought they wanted the ratio in terms of "how many tmes greater"

But the numerical difference was 1500

For the second

g(0)   =  20    and f(0)  = -40

So

g(0)  >  f(0)

1!  =  1

1! + 2!  =  1 + 2 = 3

1! + 2! + 3!  =   3 + 6  = 9

1! + 2! + 3! + 4! = 9 + 24  =  33

1 + 2! + 3!  + 4!  + 5!  = 33 + 120  =  153

Note that every successive factorial  added to his sum  will include  2 and 5  as multipliers...so every successive factorial will end in 0.....so

1! + 2!  + 3!  + 4! + 5!  .....+ 1000!   will   have a units digits sum  of 3

I'm stealing something here that I learned from heureka.....so, really....he should get the credit  !!!!

He noted that

n * n!  =

( [ n + 1 ]  - 1) * n! =

(n + 1)n! - n!  =

(n + 1)! - n!

So  we have the following

     1 * 1!           =    2!  -  1!

+   2 * 2!           =    3!  -  2!

+   3 * 3!           =    4!  -  3!

+   4 * 4!           =    5!  -  4!

+   .......

+   n * n!           =  (n + 1)!  - (n)! 

The terms in red  will "cancel"  and we will  be left with

Sum  = ( n + 1)!  - 1!  =

Sum  =  (n + 1)!  -  1

Here is the closed form, but now you have to prove it !!!

S(n) =[n + 1]! - 1

Hint: nn! = [n + 1]! - n!

The nth Lucas number  can be derived as follows

L(n)  =  Phi^n  +  (-phi)^n

Where  

Phi  =   (1 + √5) / 2 )      and   -phi  =    -2 / ( 1 + √5)

So  

L(10) =    [ 1 +  [  (  (1 + √5) / 2 )^10  +   ( -2 / ( 1 + √5) )^10 ]    = 123

Also....if you know the Fibonacci Series....

L(n)  = F(n - 1)  + F(n + 1)

So

L(10)  =  F(9)  + F(11)   =   34  + 89  =   123