All Answers 357330 Answers

1)  Rosala....I  believe  the value of  n  is 8

Reasoning

The first letter can be chosen 3 ways...either A, B or C

But....the second letter can only be chosen 2 ways.....for instance....if A  is the first letter, the second can only be chosen  2  ways, either another  A  or   C

And notice that each successive letter can only be chosen in 2 ways......for instance...if the second letter is B, the third letter can only be another B  or A

So.....we need to solve this  to find the number of additional letters, m, after the first one

 3 * 2^m  =  384       divide by 3

2^m  = 128

And m  =  7  makes this true

So....the the number of total letters  is

First one + m additional    =  1  + 7  =  8

Another way to see this is

3 * 2 * 2 * 2 * 2 * 2 * 2 * 2  =  384

It was correct! Thank you so much!

Number 3

Rosala....here's the explanation for the book's answer :

Note...if fthe leading digit cannot be 0, we have 9 choices for this

And for the next two digits, we have 10 ways to choose each

But note....that  the last digit must be even if the number is divisible by 2....so....we only have 5 choices for this  2,4,6, 8 or 0

So..... we have    9 * 10 * 10 *  5  =    4500 

Hope that helps....!!!!

After ONE year your college loan balance would be   1500 (1+.08) = 1620   if you make no paymets

Likely Lana would pay less interest on the loan as she has a HIGHER credit score !

There are  9000 numbers between 1000 and 9999 (inclusive)

  Half of these are odd numbers    A = 4500

1000 is divisible by 5  as is 9995      1 + (9995-1000)/5 = 1800 = B

A+B = 4500 + 1800 =  6300

thank you so much Melody for taking the time to answer my questions,

i'd like to start wiith the first one.

the answer that youve given to question no. 1 , thats not the answer thats present in my book. the answer to it is (x-1 )^x  - 1 , my doubt in this was that why are we subtracting one here? the (x-1)^x ca efro forula (n)^r.

2.oh Melody i feel bad to have troubled you for this problem , i was just making 2 silly mistakes. but couldnt see them,thank you for  the answer and explanation. 

3. 

#

less than      200,000,000   only containing the digits 1 and 2
1 digit     2

2 digit     2*2=2^2

3 digit      2^3

4 digit       2^4

5 digit      2^5

6 digit      2^6

7 digit      2^7

8 digit      2^8

9 digit      2^8       The biggest value digit must be 1

#

your third answer is correct btw!

i have no idea what you did here...otherwise i understand everything after that..... could you pls explain this part ? thank you for the answers! 

thank you so much Melody for taking the time to answer my questions,

i'd like to start wiith the first one.

the answer that youve given to question no. 1 , thats not the answer thats present in my book. the answer to it is (x-1 )^x  - 1 , my doubt in this was that why are we subtracting one here? the (x-1)^x ca efro forula (n)^r.

2.oh Melody i feel bad to have troubled you for this problem , i was just making 2 silly mistakes. but couldnt see them,thank you for  the answer and explanation. 

3. 

#

less than      200,000,000   only containing the digits 1 and 2
1 digit     2

2 digit     2*2=2^2

3 digit      2^3

4 digit       2^4

5 digit      2^5

6 digit      2^6

7 digit      2^7

8 digit      2^8

9 digit      2^8       The biggest value digit must be 1

#

your third answer is correct btw!

i have no idea what you did here...otherwise i understand everything after that..... could you pls explain this part ? thank you for the answers! 

Amazing, Melody! Great job!

Thanks Chris :)

Sorry to interfere here but the question is missing an important piece of information. It asks for the exact answer meaning a unique answer but it doesn't define in which quadrant the angle lies. Therefore there are an infinite number of solutions eg -5pi/6 , 7pi/6  etc. if one wants to be pedantic. In the end it's the student who suffers if they are not aware of such details. 

Actually I think it's the second quadrant perhaps you didn't see the negative sign.

The answer is 21 in.

The answer is 54

tan (2pi / 3)  = - tan (60°)   = -√3

cos ( 11 pi / 6 )  =  cos (30°)  =  √3 / 2

So....substituting, we have

-√3 +  4 (√3 / 2 )  =

-√3  + 2√3  =

√3

The number of permutations  is   6!  =  720

cos θ  =  - √ [  1  -  sin^2  θ ] =

- √ [  1  -  (4/7)^2 ]   =

- √ [  1  -  16/ 49 ]   =

- √   [ 49 - 16]  / 7  =

- √33 /  7

We take the negative root  because the cosine i negative in quadrant 2

10!  ways  =  3,628,800  ways  !!

We need to find r =   √  [  (-3)^2  + 7^2 ]  =  √ 58

So...sin  θ  =   7 / √ 58   =  7√ 58 / 58

We take the positive root because the terminal side of the angle is in quadrant 2 and the sine is positive there

Just a calculator problem

arccos (0.21)  = 77.88°

log _{a + b} m  + log _{a - b} m   -  2log _{a+ b}  m* log _{a - b} m

Note  m^2  =  a^2 - b^2   = (a + b) (a - b)

And, using the change-of-base rule, we can write

log m / log(a + b) + log  m / log (a - b)  -  2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ] )

Get a common  denominator between the first two fractions and factor out log m 

[  log( m ) [ log (a + b) + log (a - b)]  ] /  [ log(a + b)* log(a - b)] -

2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]  

{ log a + log b  =  log(a * b)  }

log (m) [ log [ (a + b)(a - b)]  / [ log(a + b)* log(a - b)] -

2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]

log (m) [ log (m^2) ]  /  [ log(a + b)* log(a - b)] -

2 [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]

{  log a^b  =  b log a }

log (m) * 2 log (m) /  [ log(a + b)* log(a - b)] -

2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]

(2 [ log (m) }^2 ( / [ log(a + b)* log(a - b)] -   (2 [ log m] ^2 ( /  [ log (a + b) * log (a - b) ] =

0

Cosine  = x / r

Cosine in quadrant  II  is negative because x is negative and r is positive

Sine  =  y / r

Sine  in quadrant II  is negative because  y, r  are both positive 

5pi/ 6  is a 30°  reference angle in the 2nd quadrant....so....

sin 5pi/6  =   sin (30°)  =  1/2

Angle of elevation   =

arctan ( 35 / 50)  =   35°

We can use  the inverse tangent ( arctan)  function here

arctan (15/8)  =  x   =  61.9°