geno3141

For the Logarithm Question:

  Assuming real numbers, square roots will be defined when the value inside the square root sign is zero or positive:

     sqrt(x + 1) is defined for all values:  x + 1 >= 0     --->     x >= -1

     sqrt(1 - x) is defined for all values:  1 - x >= 0     --->     -x >= -1     --->     x <= 1

     sqrt(x) is defined for all values:  x >= 0     --->     but a denominator can't be zero     --->     x > 0

          Putting these restrictions together:     0 < x <= 1

Polynomial Question #1:

   The degreee of a polynomial is the degree of its highest degree term.

   f(x) is quadratic   --->   its highest degree term is 2

   g(x) is cubic   --->   its highest degree terms is 3

   [ f(x) ]³  has a highest degree of  [ x² ]³  =  x⁶    

        --->   degree is 6 (with a coefficient of 1)  (but it can also have terms of degree 5, degree 4, etc.)

   [ g(x) ]²  has a highest degree of  [ x³ ]²  =  x⁶   

          --->   degree is 6 (with a coefficient of 1) (but it can also have terms of degree 5, degree 4, etc.)

   Since the coefficient of the  x⁶  term of both the [ f(x) ]3  term and the  [ g(x) ]²  term is 1, these terms will cancel      

   under subtraction, leaving a possible  x⁵  term.

--->     The maximum possible degree of the answer is 5.

Polynomial Question #2:

 If  f(x)  =  x⁷  and  g(x)  =  -x⁷,  then  f(x) + g(x)  =  x⁷ + -x⁷   =  0, which has a degree of 0, the minimum possible degree.

If  f(x)  =  x⁷  and  g(x)  =  x⁷,  then  f(x) + g(x)  =  x⁷ + x⁷   =  2x⁷,  which has a degree of 7, the maximum possible degree.

Multiplying these two answers:  0 x 7 = 0.

log_b( x · y / z )  =  log_b( x )  +  log_b( y )  -  log_b( z )

To find the HCF (highest common factor) for two numbers, factor the two numbers and use the factors that are found in both numbers (the number of times that they are found in each number):

27  =  3 x 3 x 3

45  =  3 x 3 x 5         So the HCF  =  3 x 3  =  9

18  =  2 x 3 x 3

66  =  2 x 3 x 11       So the HCF  =  2 x 3  =  6

96  =  2 x 2 x 2 x 2 x 2 x 3

44  =  2 x 2 x 11       So the HCF  =  2 x 2  =  4

sqrt( 9 / 16 )  =  sqrt( 9 ) / sqrt( 16 )  =  3/4.

33 1/3 %  is  33.33333333...  [with the threes never stopping]

                                               and  33 1/3  isn't exactly  33.33  [33.33 is only an approximation]

So:  33 1/3  +  33 1/3  +  33 1/3   is  100  [exactly].

And:  33 1/3  +  33 1/3  +  33 1/3  =  33.333...  +  33.333...  +  33.333...

                                                     =  99.999...     [with the nines never stopping]

So, why does  99.999...  equal  100?

Two different arguments (both correct);

First.  If you have two different numbers, there must be another number between them (actually many numbers between them).  So, if  99.999...  and  100  are different number,  what is a number between 99.999...   [remember that the nines never stop]?  No one can find another number between them, so they must be the same number.

Second:  an algebra proof.

  Call the number  n  to equal  99.999...  and also  find 100n  which is  9999.999...  and subtract them:

                              100n  =  9999.999999999999999...

                                    n  =      99.999999999999999...

Subtracting:             99n  =  9900.000000000000000...

Dividing by 99:             n  =  100.0000000000000000...  or, more simply,  100

Therefore,       99.999...  =  100

Let  X  =  amount of water

Final Amount  =  X + 30

Percentage of sulfuric acid in water  =  0%

Final Amount x Final Percentage  =  (First Amount x First Percentage) + (Second Amount x Second Percentage)

        (X + 30) x (0.23)                    =  (                30 x 0.36                   ) + (                       X x 0.00                         )

           0.23X + 6.90                       =                      10.80                        +                            0.00

                                0.23X + 6.90  =  10.80

                                         0.23X  =  3.90

                                                 X  =  16.96 ml

Let  W  =  width of the plot.

Let  L  =  length of the plot.

L  =  2W - 3

Total perimeter =  L + W + L + W

                 510  =  (2W - 3) + (W) + (2W - 3) + (W)

                  510  =  6W - 6

                  516  =  6W

                    86  =  W

L  =  2W - 3     --->     L  =  2(86) - 3  =  169

4{ 8z - [9 - 7(7z - 5)] }

Working from the inside out:  -7(7z - 5)  =  -49z + 35

4{ 8z - [9 - 49z + 35] }

Simplifying:  9 - 49z + 35  =  44 - 49z

4{ 8z - [44 - 49z] }

Again, working from the inside out:  8z - [44 - 49z]  =  8z - 44 + 49z  =  57z - 44

4{ 57z - 44 }

228z - 176

x  =  2^1/3 + 2^2/3     --->     x³ - 6x - 6  =  0

First, let's evaluate:  [ 2^1/3 + 2^2/3 ]³

 Notice that  [ a + b ]³  =  a³ + 3a²b + 3ab² + b³

The x³ part:     [ 2^1/3 + 2^2/3 ]³  =  [2^1/3]³ + 3·[2^1/3]²·[2^2/3] + 3·[2^1/3]·[2^2/3]² + [2^2/3]³

                                              =  2^3/3 + 3·[2^2/3]·[2^2/3] + 3·[2^1/3]·[2^4/3] + 2^6/3

                                              =  2 + 3·2^4/3 + 3·2^5/3 + 4

                                              =  6 + 3·2·2^1/3 + 3·2·2^2/3 

                                              =  6 + 6·2^1/3 + 6·2^2/3 

The 6x part:    6·[ 2^1/3 + 2^2/3 ]  =  6·2^1/3 + 6·2^2/3 

Thus:  x³ - 6x - 6  =  [ 6 + 6·2^1/3 + 6·2^2/3  ] - [ 6·2^1/3 + 6·2^2/3  ] - 6

                             =  6 + 6·2^1/3 + 6·2^2/3 - 6·2^1/3 - 6·2^2/3 - 6

                             =  0

Problem:  2.836^x  =  286 330 760 848.37

Use logarithms; you can use either common logarithms (log) or natural logarithms (ln).

--->         log( 2.836^x )  =  log( 286 330 760 848.37 )

--->     x · log( 2.836 )  =  log( 286 330 760 848.37 )

--->                           x  =  log( 286 330 760 848.37 ) / log( 2.836 )

--->                           x  =  25.30751144  (approx)

In step 2, any exponent within a log (or ln) expression can come out as a multiplier.