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1.

Find the indicated sum. Show your work.

k = 1 (-1)^k (k + 11) = (-1)^(1) (1 + 11)= -1*(12) = -12

k = 2 (-1)^k (k + 11) = (-1)^(2) (2 + 11)= 1*(13) = 13

k = 3 (-1)^k (k + 11) = (-1)^(3) (3 + 11)= -1*(14) = -14

k = 4 (-1)^k (k + 11) = (-1)^(4) (4 + 11)= 1*(15) = 15

(-12)+(13)+(-14)+(15)=2

2.

Locate the foci of the ellipse. Show your work.

X^2=(x-h)^2 then h=0

Y^2=(x-k)^2 then k=0

The centre is (00)

X^2/36+y^2/11=1

When x=0 y^2/11=1; y=0

When y=0x=0

X^2/36=1;x=0

11+c^2=36

C=5

Foci (50) and (-50)

3.

Solve the system by the substitution method. Show your work.

2y – x =5

x2 + y2 -25 = 0

x:

2y – x = 5

2y – 5 = x

so x = 2y – 5

-Plug this into 2nd equation:

(2y – 5) + y – 25 = 0

-Use FOIL to solve the (2y – 5) part:

(2y – 5)(2y – 5)

4y – 10y – 10y + 25

4y – 20y + 25

So :

4y – 20y + 25 + y – 25 = 0

Which can be simplified to:

4y + y – 20y = 0

4y + y – 20y = 0

y(4y + y – 20) = 0

So because of the 0 multiplication rule

y=0

x= -5 (plug in y=0 to original equations:

2y – x = 5

2(0) – x = 5 so x= -5)

(-50)

Y(4y+y-20)=0

So y=0 or 4y+y-20=0

5y-20=0

Y=4

X=2y-5 when y=4

X=8-5=3

(-50) (34)

4.

Graph the function. Then use your graph to find the indicated limit. You do not have to provide the graph

f(x) =5x- 3 f(x)

22

5.

Use Gaussian elimination to find the complete solution to the system of equations or state that none exists. Show your work.

4x – y + 3z = 12

x + 4y + 6z = -32

5x + 3y + 9z = 20

6.

Solve the system of equations using matrices. Use Gaussian elimination with back-substitution.

x + y +z = -5

x – y+ 3z = -1

4x + y +z = -2

X=1/3 y=-(11/3)z=-(5/3)

7. A woman works out by running and swimming.When she runs she burns 7 calories per minute.When she swims she burns8 calories per minute.She wants to burn at least336 calories in her workout.Write an inequality that describes the situation. Letx represent the number of minutes running and y the number of minutes swimming.Because x and y must be positive limit the boarders to quadrant I only.

7x+8y>=336

Short Answer Questions: Type your answer below each question. Show your work.

8.

A statement Sn about the positive integers is given. Write statements S1 S2 and S3 and show that each of these statements is true. Show your work.

Sn: 12 + 42 + 72 + . . . + (3n – 2)2 =

S1=1(6*1^2-3(1)-1)/2=1

S2=1^2+4^2=17

S31^2+4^2+7^2=66

9.

A statement Sn about the positive integers is given. Write statements Skand Sk+1 simplifying Sk+1 completely. Show your work.

Sn: 1 2 + 2 3 + 3 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3

10.

Joely’s Tea Shop a store that specializes in tea blends has available 45 pounds of A grade tea and 70 pounds of B grade tea. These will be blended into 1 pound packages as follows: A breakfast blend that contains one third of a pound of A grade tea and two thirds ofa pound of B grade tea and an afternoon tea that contains one half pound of A grade tea and one half pound of B grade tea. If Joely makes a profit of $1.50 on each pound of the breakfast blend and $2.00 profit on each pound of the afternoon blend how many pounds of each blend should she make to maximize profits? What is the maximum profit?

Grade A = 45 lbs

Grade B = 70 lbs

Total weight = A+B = 45+70 = 115

Blend Br = 1/3A + 2/3B

Blend Af = 1/2A + 1/2B

Profit = 1.50*lbs Blend Br + 2.00*lbs Blend Af

Profit = 1.50*(1/3A + 2/3B)+200(1/2A + 1/2B) =

45 pounds of A and 70 pounds of B yields max 24A + 72B = 98 Pounds Max Br

max 38A + 76B = 114 Pounds Max Af

The rate of profit for A is .50 per pound in BR and 1.00 per Pound in AF

For B rate of profit for B is 1.00 for Br and 1.00 for Af.

Let X = # pounds in A

Let Y = # pounds in B

115 = A + B

A = 115-B

Let q = percent of Br

Let r = percent of Af

Let s = pounds of A in Br then 45-s = # pounds A in Af

Let t = pounds of B in Br and 70-t= # pounds in Af

.5s + 1.00*(45-s) + 1.00(t) + 1.00(70-t) = p

1.5s + 2.00t = p

If s = 45 pounds then 70 = 70/45 = 14/9s pounds

.5s + 1(45-s) + 1(14/9s) + 1(70-14/9s) = p

.5*45s + (45(1 – s)) + 14/9*45s + (45(1 – 14/9s) =p

22.5s + 45 – 45s + 70s + 35 – 70s = p

22.5s +70 – 115s = p

p = 137.5s + 70

p’ = 137.5

22.5 A and 45 B for Br and 23.75 A and 23.75 B for Af.

Max profit is $137.5

11

Your computer supply store sells two types of laser printers. The first type A has a cost of $86 and you make a $45 profit on each one. The second type B has a cost of $130 and you make a $35 profit on each one. You expect to sell at least 100 laser printers this month and you need to make at least $3850 profit on them. How many of what type of printer should you order if you want to minimize your cost?

Let x represent the number of A printers

Let y represent the number of B printers

Minimize cost = 86x + 130y

subject to

Total printers equn: x + y 100

Total profit equn: 45x + 35y 3850

x 0 y 0

x and y must be whole numbers.

The vertices of the feasible region are: (0 100) (100 0) and (35 65)

If x = 35 and y = 65 the cost is 11460 and profit is 3850

if x = 100 and y = 0 the cost is 8600 and profit is 4500

If x = 0 and y = 100 the cost is 13000 and profit is 3500

The best result is x = 100 and y = 0

12

A statement Sn about the positive integers is given. Write statements S1 S2 and S3 and show that each of these statements is true. Show your work.

Sn: 2 +5 +8 + . . . + ( 3n -1) = n(1 + 3n)/2

13

Use mathematical induction to prove that the statement is true for every positive integer n. Show your work.

2 is a factor of n2 – n + 2

14

A statement Sn about the positive integers is given. Write statements S1 S2 and S3 and show that each of these statements is true. Show your work.

Sn: 2 is a factor of n2 +7n

15

(i.) f(x)

(ii.) f(x)

(iii.) What can you conclude about f(x)? How is this shown by the graph?

(iv.) What aspect of costs of renting a car causes the graph to jump vertically by the same amount at its discontinuities?

16

Use mathematical induction to prove that the statement is true for every positive integer n.

8+ 16+ 24 + . . .+ 8n =4n(n + 1)

8(1)=4(1)(1+1)

8=8

1st condition is true

8+16+24++8k+8(k+1)=4k(k+1)+8(k+1)1

8+16+24+8k=4k(k+1)

=4k(k+1)+8(k+1)

=4k^2+4k+8k+8

=4k^2+12k+8

=4(k^2+3k+2

=4(k+1)((k+1)+1)

17

The following piecewise function gives the tax owed T(x) by a single taxpayer on a taxable income of x dollars.

T(x) =

(i) Determine whether T is continuous at 6061. Yes the function is continuous at 6061

(ii) Determine whether T is continuous at32473. T is continuous at 32473

(iii) If T had discontinuities use one ofthese discontinuities to describe a situation where it might be advantageous to earn less money in taxable income. T is continuous at every point it has not discontinuities so earning less money for saving tax isnt good idea

18

A statement Sn about the positive integers is given. Write statements Sk and Sk+1 simplifying Sk+1 completely.

Sn: 1 + 4 + 7 + . . . + (3n – 2) = n(3n – 1)/2

19

An artist is creating a mosaic that cannot be larger than the space allotted which is 4 feet tall and 6 feet wide. The mosaic must be at least 3 feet tall and 5 feet wide. The tiles in the mosaic have words written on them and the artist wants the words to all be horizontal in the final mosaic. The word tiles come in two sizes: The smaller tiles are 4 inches tall and 4 inches wide while the large tiles are 6 inches tall and 12 inches wide. If the small tiles cost $3.50 each and the larger tiles cost $4.50 each how many of each should be used to minimize the cost? What is the minimum cost?

0 small tile 6 large tiles and a minimum cost of 27$

20

The Fiedler family has up to $130000 to invest. They decide that they want to have at least $40000 invested in stable bonds yielding 5.5% and that no more than $60000 should be invested in more volatile bonds yielding 11%. How much should they invest in each type of bond to maximize income if the amount in the stable bond should not exceed the amount in the more volatile bond? What is the maximum income?

X= amount of money invested in the stable bonds at 5.5%

Y= amount of money invested in the volatile bonds at 11%

Up to 130000 invested which means the investment can not be larger than 130000

X+y<=130000
Stable bonds= 130000-60000=70000
At least 40000 invested in stable bonds yielding 5.5%
x>=40000

no more than 60000 should be invested in more volatile bonds yielding 11%

y<=60000
60000 in the stable bonds and 60000 in volatile bonds
Maximum income $9900