All Answers 358239 Answers

2 obviously unless you get funky and solve it using -2/3

Thanx Melody !  

Sequence A:  a_n = 6^(n-1)  or  a_n = 2^(n-1)*3^(n-1)

Sequence B:  b_n = 50*3^(n-1)

For A to be larger than B:   2^(n-1)*3^(n-1)  >  50*3^(n-1)   

or   2^(n-1)  >  50  

The smallest value of n for which 2^(n-1)  >  50  is n = 7  (as 2⁵ = 32 and 2⁶ = 64) 

Hence after 7 terms A is larger than B.

A sequence is 6^(n-1)     B sequence is 50(3^(n-1))

Set them equal and solve for n

6^(n-1) = 50(3 ^(n-1))

 Take log of both sides

(n-1)Log 6 = Log50 +(n-1)log3 

(n-1)(log6-log3) = log 50

n-1 = log50 /(log6-log3)

n = log50/(log6-log3) + 1

n= 6.64    This is where they would EQUAL,,,,the nex integer is where A>B           = 7

the integer after which A>B is 6

How come my birthday wasnt recognized?

Cmon mate give me something to work with.

1. (5/8)x - (3/8)x = 2         Write the problem out

2. (2/8)x = 2                     Combine like terms

3. 2x = 16                         Multiply both sides by 8 to break the fraction

4. x = 8                             Divide both sides by 2

A quark. There's three quarks in each proton/neutron

£5.58

Hi ElectricPavlov and asker.  Here is more of a look if you are interested :)

y = 4 cos 1/3 ( theta + 45 degrees ) - 3]

\(y = 4 cos[\frac{ 1}{3} (\theta + 45 ) ]- 3\)

I like to rewrite this as

\(y+3 = 4 cos[\frac{ 1}{3} (\theta + 45 ) ]\)

1)  amplitude=4

2) Centre of wave   y+3=0      y=-3

           (there is probably a better name for this but I don't know it)  

3) \(Period = \frac{360}{\frac{1}{3}} = 360\div\frac{1}{3} = 360*3 =  1080\; degrees\)

4) Phase shift    

 \(\theta+45=0 \qquad \theta=-45\\ \mbox{That is 45 degrees in the negative direction}\)     

Here is a you tube clip if you want to watch it :)

Yes I am with you ElectricPavlov, we need a theta,

The phase shift is actually   -45 degrees

I think of it like this

\(\theta+45=0\\ \theta=-45\\ \)

Maybe that "logic' won't work for you but it works for me  :)

Here is a graph.

Just add the f3 components in a similar manner.

Pi symbol = π 

π = 3.141592653....

Universally Shortened = 3.14

Remember the shortened numbers like this:       P  I  E

                                                                            4  1. 3    TURN IT AROUND

                                                                            3.  1 4

How about the f3?

Try them this way:

x-component of resultant, fx = f1*cos(theta1) + f2*cos(theta2)

y-component of rsultant,  fy = f1*sin(theta1) + f2*sin(theta2)

magnitude of resultant  f = sqrt(f1^2 + f2^2)

angle of resultant to positive x-axis  theta = tan^-1(fy/fx)

x^5 terms can only arise from the following combinations:  

2x*x^2*x^2    coeff → 2

3x^2*2x*x^2  coeff → 6

3x^2*x^2*x   coeff → 3

So overall coefficient of x^5 tterm is 2 + 6 + 3 → 11  (solution A in your list).

I should have noted that A in your list is the right answer.

Write it as (2 + 3x)^9

Expand:

(2 + 3x)^9 = 2^9 + 9*2^8*(3x) + (9*8/2!)*2^7(3x)^2 + (9*8*7/3!)*2^6*(3x)^3 + (9*8*7*6/4!)*2^5*(3x)^4 + ....

So coefficient of x^4 is (9*8*7*6/4!)*2^5*3^4 → 326592

.

The analysis below should help you to work out the angles between them:

.

Subatomic particle, out of those the electron is the smallest.

Lots of people ask question.

Some of the regular answerers are already mathematians or near/at university level. They mostly come to answer, not to ask  :)

11.8 x 10^9

=1.18×10×10^9

=1.18×10^10

I'll hazard a guess...first of all I will guess that  0+45  is really 'theta' +45    and  I'll guess that 45 would be the phase shift.

The 1/3 in the equation lends me to believe the period will be 360 x 3  ..... these are all guesses....educate me...THANX !