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MATH 154 PRACTICE MIDTERM III

Please work out each of the given problems.  Credit will be based on the steps that you show towards the final answer.  Do all your work and give all your answers on you own sheet of paper.  Show your work.

Problem 1 (20 Points)  Solve the following system of nonlinear equations:

        x2 + y2  =  25

        3x + 5y  =  15

  Solution

We first solve the second equation for x by dividing by subtracting 5y and dividing by 3

        3x  =  15 - 5y

        x  =  5 - 5/3 y

Now substitute into the first equation to get

        (5 - 5/3 y)2 + y2  =  25

        25  -  50/3 y + 25/9 y2 + y2  =  25        FOIL

        -50/3 y + 25/9 y2 + y2  =  0        Subtracting 25 from both sides

        -150y + 25y2 + 9y2  =  0        Multiplying both sided by 9

         34y2 - 150y  =  0        Combining like terms

          2y(17y - 75)  =  0    Factoring out the GCF

           y  =  0        or      y  =  75/17  =  4.41

Now plug these solutions back into the equation

           x  =  5 - 5/3 y

to get

           x  =  5 - 5/3 (0)        or        x  =  5 - 5/3 (75/17)

           x  =  5        or        x  =  5 - 125/17  =  -40/17  =  2.35

Our two solutions are (5,0) and (2.35,4.41)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 2

A. (8 Points)  Find the first four terms of the sequence whose general term is given. Then find the 100th term. 

        an  =  n - 1/n

Solution

We have

        a1  =  1 - 1/1  =  0

        a2  =  2 - 1/2  =  3/2

        a3  =  3 - 1/3  =  8/3

        a4  =  4 - 1/4  =  15/4

        a100  =  100 - 1/100  =  9999/100

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B. (8 Points)  Expand and simplify

       

Solution

We plug in 2, 3, 4, and 5 to get

        [2(2) + 1] + [2(3) + 1] + [2(4) + 1] + [2(5) + 1] 

        =  5 + 7 + 9 + 11  =  32

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C. (8 Points)  Write the series in summation notation

        1       2       3       4
            +       +       +       + ...
        2       3        4       5                       

Solution

The numerator is just n and the denominator is n + 1.  Where n starts at 1 and continues to infinity.  In summation notation, this becomes

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 3 

A)     (11 Points)  Find the indicated unknown

        8, 13, 18, 23, ..., 88            n  =  ?

Solution

This is an arithmetic sequence since 

        13 - 8  = 18 - 3  =  23 - 18  =  5

The common difference is d  =  5.

The first term is 8, so the general formula is

        an  =  8 + 5(n - 1)

The last term is 88, so we set

        88  =  8 + 5(n - 1)    

        80  =  5(n - 1)    Subtract 8

        16  =  n - 1        divide by 5

        n  =  17        add 1

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B)    (11 Points)  Find the sum of the given series

       

Solution

This is an arithmetic sequence with 

        d  =  3       a1  =  3        and         n  =  54

We use the sum of an arithmetic sequence formula to find the 54th term.

        a54  =  3 + 3(54 - 1)  =  162

The sum is 

        n/2 (a1 + an)

        54/2 (3 + 162)  =  4455

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 4 

A. (11 Points)  Find the indicated unknown

        a1  =  5,   a5  =  0.008        r  =  ?

Solution

This is a geometric sequence with 

        an  =  a1rn-1

We have

        0.008  =  5r5-1

        0.0016  =  r4        Dividing by 5

         r  =  (0.0016)1/4  =  0.2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B. (11 Points)  A rubber ball is dropped on a hard surface and bounces to a height of 200 ft.  On each rebound it bounces 95% as high as on the previous bounce.  How high does the ball bounce on the 20th bounce?

Solution

We first write the first few terms of the sequence, which represents the height of the nth bounce.

        a1  =  200        a2  =  .95(200)      a3  =  .95(.95(200))  =  .952(200)

We see that this is a geometric sequence with 

        an  =  200(.95)n-1 

We want the height of the 20th bounce.  We have

        a20  =  200(.95)19  =  75.5

The ball bounces about 75.5 feet on the 20th bounce

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 5 

A. (11 Points)  Find the sum of the infinite series if it exists.

              1       1         1       
       1+       +        +         + ...
              3       9         27                       

Solution

This is a geometric sequence with first term 1 and common ratio r  =  1/3. We have

                          a1                   1                        1              3
             S  =                   =                    =                    =         
                         1 - r               1 - 1/3                2/3            2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B. (11 Points)  Write the repeating decimal as the ratio of two integers.

        0.18   

 

Solution

We write

        0.18   =  .18 + .0018 + .000018 + .00000018 +...

This is an infinite geometric sequence with 

        a1  =  .18  =  18/100    and     r  =  .01  =  1/100

We have 

                    18/100            
        S  =                                
                   1 - 1/100

                    18            
         =                           Multiply top and bottom by 100     
                  100 - 1

                 18               2
          =              =            
                 99              11

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 6  (20 Points)

Expand the binomial using the binomial theorem

        (3m - 2)5

Solution

We use Pascal's Triangle

            1
          1   1
        1    2    1
     1    3     3     1
   1    4     6     4    1
1    5    10   10    5    1

The bottom line represents the coefficients.  We have

        (3m - 2)5  

        =  1(3m)5(-2)0 + 5(3m)4(-2)1 + 10(3m)3(-2)2 + 10(3m)2(-2)3 + 5(3m)1(-2)4 + 1(3m)0(-2)5

        =  243m5 + 5(81m4)(-2) + 10(27m3)(4) + 10(9m2)(-8) + 5(3m)(16) + 1(-32)

        =  243m5 - 810m4 + 1080m3 - 720m2 + 240m - 32

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 7  (20 Points)

One muffin, two pies, and three cakes cost $23.  One Muffin, three pies , and two cakes cost $21.  One muffin, four pies, and five cakes cost $39.  Find the cost of each.

Solution

Let

        x  =  the cost per muffin

        y  =  the cost per pie

        z  =  the cost per cake

We get three equations:

        x + 2y + 3z  =  23

        x + 3y + 2z  =  21

        x + 4y + 5z  =  39

Subtracting the second equation from the first gives

        -y + z  =  2

Subtracting the second equation from the third gives

        y + 3z  =  18

Now add the two equations above to get

        4z  =  20

        z  =  5    dividing by 4

Plugging z  =  5 back into the equation above to get

        y + 3(5)  =  18

        y  =  3        subtracting by 15

Now plug in y  =  3 and z  =  5 into the first original equation to get

        x + 2(3) + 3(5)  =  23

        x + 21  =  23            6 + 15  =  23

        x  =  2                       Subtracting 2 from both sides

Now plug in (2,3,5) into the second and third equations to check.

        2 + 3(3) + 2(5)  =  21

        2 + 4(3) + 5(5)  =  39

We can conclude that the muffins cost $2, the pies cost $3, and the cake costs $5.