All Answers 357307 Answers

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

what do we do to help you?

Thanks so much, both of you!

what do we find?

Hmmmm.... what should I get for a bulemic girl that doesn't like flowers, chocolate, spas, jewelry, perfume, cars or gift cards?

   LOL       JK       

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)

no problem 

Happy Valentines day and thanks for your support and help.

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

:D

thumbs up

I am alright thanks.

living the dream

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