geno3141

If 1 + 6 + 11 + ... + 96 + 101  =  n (mod 15), what is the value of n?

I believe that Guest's answer of 6 is correct; this is my slow, step-by-step way of finding that answer.

I plan to find the sum of the terms, divide that answer by 15, and the remainder will be n.

The series is an arithmetic series.

The formula for the sum of an arithmetic series is:  Sum  =  n(f + l) / 2

where  n = number of terms,  f = first term,  and  l = last term.

f = 1  and  l = 101

but I still need to find the number of terms.

The formula for the n^th term of an arithmetic sequence is:  t_n =  t₁  + (n - 1)d

t₁ = first term = f = 1

t_n = last term = l = 101

d = common difference = 5

Substituting these into the formula:  t_n =  t₁  + (n - 1)d     --->     101  =  1 + (n - 1)·5

                                                                    Solving:                   100  =  (n - 1)·5

                                                                                                      20  =  n - 1

                                                                                                         n = 21  (there are 21 terms in the series)

Now, solving for the sum:  Sum  =  n(f + l)/2

                                          Sum  =  21(1 + 101) / 2

                                          Sum  =  1071

Since we're working with mod 15, let's divide the sum by 15; the remainder will be our answer.

     1071 / 15  =  71, and the remainder is 6.

Therefore, our answer is 6.

If you know that  5x = 1s  and  10s = 1b,

then you can multiply the first equation to get  50x = 10s,

and you can tie this equation to the second equation to get:  50x = 10s = 1b.

Now, if you want 50b:

Multiply  50x = 10s = 1b  by 50  to get  2500x = 50s = 50b,

so  50b  = 2500x.

The absolute maximum is the largest value for the function for all the values of x from negative infinity to positive infinity.

If you graph  f(x)  =  [x² - 4x + 1 ] / [ x² + 1 ]  you will find that the highest value for f(x) occurs at (-1, 3).

Now, the problem is to get the maximum value to occur at (-1, 5).

If you multiply f(x) by 3 [to create  g(x)  =  3f(x)], the maximum value occrs at (-1, 9).

Now, subtract 4 from this function, creating  g(x)  =  3f(x) - 4, the maximum value occurs at (-1, 5).

Thus, the value of b is 3.

The solution depend upon whether you meant (1/4)^x + 2  =  16  or  (1/4)^{x  + 2}  =  16 ...

      (1/4)^x + 2  =  16     

            (1/4)^x  =  14                      Since the unknown is an exponent, find the log of both sides:

   log( (1/4)^x )  =  log(14)               A property of logs allows the exponent to come out as a multiple:

    x · log(1/4)  =  log(14)

                    x  =  log(14) / log(1/4)     --->     x = -1.9036...

(1/4)^{x +  2}  =  16                             Since 1/4  =  4^-1:

 (4^-1)^{x + 2}  =  16                             Since you have an exponent to an exponent, multiply the exponents:

      4^{-x - 2}  =  4²                              Since the bases are equal, the exponents must be equal:

       -x - 2  =  2

            -x  =  4

             x  =  -4

Part a is not an identity.

If you try to establish:  sin(t) / ( 1 + cos(t) )   =   csc(t)

--->     sin(x) / ( 1 + cos(t) )  ·  [ ( 1 - cos(t) ) / ( 1 - cos(t) )             

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ ( 1 + cos(t) ) · ( 1 - cos(t) ) ]

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ 1 - cos²(t) ]                       since  1 - cos²(t) =  sin²(t) --->

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ sin²(t)  ]       

--->     ( 1 - cos(t) ) / sin(t)

--->     1 / sin(t) - cos(t) / sin(t)

--->     csc(t) - cot(t)                                        not just csc(t) 

Part b:      ( sec²(t) - 1 ) / sec²(t)  =  sin²(t)

--->     ( sec²(t) - 1 ) / sec²(t)

--->     sec²(t) / sec²(t)  -  1 / sec²(t)

--->     1 - cos²(t)

--->     sin²(t)

Part c:     Write 1 / [ 1 + sin(x) ]  without fractions:

--->     1 / [ 1 + sin(x) ]  ·  [ ( 1 - sin(t) ) / ( 1 - sin(t) ) ]

--->     [ 1 · ( 1 - sin(t) ) ] / [ ( 1 + sin(t) ) · ( 1 - sin(t) ) ]

--->     [ 1 - sin(t) ] / [ 1 - sin²(t) ]

--->     [ 1 - sin(t) ] / [ cos²(t) ]

--->     1 / cos²(t)  -  sin(t) / cos²(t)

--->     sec²(t)  -  sin(t) / cos(t) · 1 / cos(t)

--->     sec²(t)  - tan(t) ·sec(t)

Problem:  x² - 5x + 6  =  2(x - 4)²  has answers  A  and  B.   Find the value of AB - A²

  x² - 5x + 6  =  2(x - 4)²  

 x² - 5x + 6  =  2(x - 4)(x - 4)

 x² - 5x + 6  =  2(x² - 8x + 16)

 x² - 5x + 6  =  2x² - 16x + 32

Subtract x² from both sides:     - 5x + 6  =  x² - 16x + 32

Add 5x to both sides:                          6  =  x² - 11x + 32

Subtract 6 frm both sides:                  0  =  x² - 11x + 26

Use the quadratic equation with a = 1, b = -11, and c = 26:

      A  =  [ -(-11) + sqrt( (11)² - 4(1)(26) ) ] / [ 2(1) ]           B  =  [ -(-11) - sqrt( (11)² - 4(1)(26) ) ] / [ 2(1) ]

      A  =  [11 + sqrt( 121 - 104 ) ] / [2]                                B  =  [11 + sqrt( 121 - 104 ) ] / [2]

      A  =  [ 11 + sqrt( 17 ) ] / [ 2 ]                                         B  =  [ 11 + sqrt( 17 ) ] / [ 2 ]

AB - A²      --->     [ 11 + sqrt( 17 ) ] / [ 2 ]  ·  [ 11 + sqrt( 17 ) ] / [ 2 ]   -   { [ 11 + sqrt( 17 ) ] / [ 2 ] }²

                 --->     [ 121 - 17 ] / 4     -     [ 121 + 11sqrt(17) + 11sqrt(17) + 17 ] / 4

                 --->          104 / 4      -     [ 138 + 22sqrt(17) ] / 4

                 --->     [ - 34 - 22sqrt(17) ] 4

What is the sum of a finite geometric series whose first term is 5, common ratio is 3, and n^th term is 1215?

The n^th term of a geometric series can be found by using the formula:  a_n  =  a₁ · r^{n - 1}.

For this problem,  a_n = 1215,  a₁ = 5, and r = 3   --->   1215  =  5 · 3^{n - 1}

     divide both sides by 5:                                               243  =  3^{n - 1}

     take the log of both sides:                                   log(243)  =  log(3^{n - 1})

     simplify:                                                               log(243)  =  (n - 1) · log(3)

     divide both sides by log(3):                     log(243) / log(3)  =  n - 1

     simplify:                                                                           5  =  n - 1

     add 1 to both sides:                                                         n  =  6

To find the sum, use this formula:     Sum  =  a(1 - rⁿ) / (1 - r)

                                                          Sum  =  5(1 - 3⁶) / (1 - 3)

                                                          Sum  =  5(-728) / (-2)

                                                          Sum  =  -3640 / -2

                                                          Sum  =  1820 

Standard deviation is a measure of a set of data, measuring how close the data group around their central value (the mean).

It does not apply to a single number (18912/11).

1)  Using a 3 x 3 ft square to estimate a crowd in a 100 x 150 ft field. What would the equation/formula be?

          Since a 3 ft x 3 ft square contains 9 square feet and a 100 ft x 150 ft field contains 15,000 square feet:

          let  x  be the number of persons in a 3 ft x 3 ft square and  y  be the number of persons in a 100 ft x 150 ft field

          --->   y  =  (15 000 / 9) · x

2)  Using a 10 x 10 ft square to estimate a crowd in a 250 x 250 ft field. What would the equation/formula be?

         Since a 10 ft x 10 ft square contains 100 square feet and a 250 ft x 250 ft field contains 62,500 square feet:

          let  x  be the number of persons in a 10 ft x 10 ft square and  y  be the number of persons in a 250 ft x 250 ft field

          --->   y  =  (62 500 / 100) · x                      --->   y  =  625x

Solve:     (3x + 2)⁴  =  232

First:  find the fourth root of both sides:     3x + 2  =  232^1/4                            3x + 2  =  3.90276 ...

Second:  subtract 2 from both sides:               3x  =  232^1/4 - 2                      3x  =  1.90276 ...

Third:  divide both sides by 3:                            x  =  (232^1/4 - 2) / 3                x  =  0.63425,,,