# N-3/8=6

Tuesday, September 7th, 2010

# If Utility Function is Expressed as U(x,y) = x^0.5 y^0.5 What is the Marginal Utility at Point (64,25) and (49,36)? Treat Y as a Constant. Would the Answer be MUx(x,y) = y. MUx(64,25) = 5 ? and MUx(49,36) = 6?

Tuesday, September 7th, 2010

### Solution:

MUx(x,y) =     Derivative of U(x,y), treating y constant.

MUx(64,25) =

MUx(49,36) =

# Is 288/1155 in Lowest Terms

Tuesday, September 7th, 2010

### Solution:

1. First, factorize the numerator using prime numbers: 288 = 2*12*12 = 2*3*4*3*4=2*2*2*2*2*3*
2. Second, factorize the denominator: 1155 = 5*231=5*7*33 = 5*7*3*11
3. Notice that the numerator and denominator have only one factor in common (3).
4. Restate the ratio after removing the common factor
1. New numerator = 2*2*2*2*2*3 = 96
2. New denominator = 5*7*11 = 385
3. Same ratio in lowest terms:

# What Experiment Have 3 Types of Variables?

Tuesday, September 7th, 2010

### Solution:

A scientific experiment has three types of variables: independent, dependent and controlled. Please click on this link to get details: http://www.sciencebuddies.org/mentoring/project_variables.shtml.

# Factor These: 2w^2-5w-10

Tuesday, September 7th, 2010

### Solution:

We can factor the expression  as follows:

To figure this out, we used the quadratic formula (see http://en.wikipedia.org/wiki/Quadratic_equation):

w =

We recognized that y=  is a quadratic equation in the form of y=  where a=2, b=-5 and c=-10.

To confirm that our calculation are right, we decided to plot the graph of y = by plugging in various values of w.

As you can see, there is a root at approximately -1.3 and another root at approximately +3.8. So we know approximately what the values of a and b are. We could use a calculator to get closer and closer to the exact values of a and b. For example, if we try w=3.8, we get a value of y = -0.12. If we try a slightly different value, we can see if we’re getting closer to zero or farther away: w=3.81 produces a value of y=-0.018, so we know that 3.81 is a better approximation than 3.8. If we keep calculating closer approximations of the root, we will arrive at the value of b that we calculated using the quadratic formula:

# Use Mathematical Induction to Prove That the Statements are true for Every Positive Integer n. 1*3+2*4+3*5+â‹¯+n(n+2)=n(n+1)(2n+7)/6

Monday, September 6th, 2010

### Solution:

First, let’s make sure we understand what we mean by “mathematical induction”: “Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true of all natural numbers. It is done by proving that the first statement in the infinite sequence of statements is true, and then proving that if any one statement in the infinite sequence of statements is true, then so is the next one.” (Wikipedia) We have pasted below another more detailed explanation of how to create a proof using mathematical induction (see “Appendix” below).

Let’s now take the equation you provided: 1*3+2*4+3*5+…+n(n+2)=

We can show that this is true for n=1:         1*3 = = = 3

Now let us assume that the statement is true for n = k. If it is, then we will prove that it has to be true for n=k+1:

WTS è =       +

=

=

=

=

=       QED

In the above proof, WTS means “want to show” and QED means “quod erat demonstratum” (“which was to be demonstrated”).

## Appendix

Here’s another more detailed explanation of how to create a proof using mathematical induction:

# A Jeweler Needs to mix an Alloy with 16% Gold Content & an Alloy with a 28% Gold Content to Obtain 32oz. of a new Alloy with a 25% Gold Content. How Many oz. of Each of the Original Alloys Must be Used?

Monday, September 6th, 2010

### Solution:

• A = number of ounces of the first alloy
• B = number of ounces of the second all
• A+B=32
• 0.16 A +0.28 B = 32 x 0.25
• Since A + B = 32, we know that A = 32 – B. You can then re-write the equation above as 0.16 (32-B) + 0.28 B = 32 x 0.25
• This can be simplified to 5.12- 0.16 B + 0.28 B =8
• Or, 0.12 B = 2.88
• Or B = 24
• So A = 8

# Solve the Triangle. Round Angle Measures to the Nearest Minute & Side Measures to the Nearest Tenth. a=11.4, b=13.7, c=12.2

Monday, September 6th, 2010

### Solution:

These are the formulas used to solve triangles:

1. The sum of the internal angles equals 180o …A + B + C = 180o

2. The ‘sine rule’ …

3. The ‘cosine rule’ …

a² = b² + c² – 2bc cosA

or

b² = a² + c² – 2ac cosB

or

c² = b² + a² – 2ba cosC

In the problem that you gave us, we know the length of the three sides of the triangle:

a = 11.4

b = 13.7

c = 12.2

When no angles are known, the cosine rule is the only option, so 11.42 =13.72 +12.22 – 2*13.7*12.2*cosA….Therefore cosA = 0.617955.

You can now use a calculator or a table to find the value of A. You should get A = 0.9047 radians = 51.833 degrees = 51 degrees and 50 minutes.

Use the sine rule to find one of the remaining angles.

=

So sin(B) =  = 0.944835

You can now use a calculator or a table to find the value of B. You should get B = 1.237 radians = 70.8801 degrees = 70 degrees and 53 minutes.

Finding the third angle is easy, since we know that A + B +C =180 degrees. C = 180 – 51.833 – 70.880 = 57.287 degrees = 57 degrees and 17 minutes.

# What is the Largest Amount of Postage That Cannot be Made Using These Stamp Denominations: 17 Cents, 9 Cents & 5 Cents?

Monday, September 6th, 2010

### Solution:

We’re not sure we understand the question, because if you have enough stamps (of any denomination) you can cross any amount. However, you said that the total value (let’s call is “q”) has to be less than a million dollars (e.g. \$999,999.99), so let’s work backwards.  \$999,999.99 can be obtained as follows:

1. 5,882,350 stamps of 17 cent denomination
2. 1 stamp of 9 cent denomination
3. 8 stamps of 5 cent denomination

So we have shown the largest amount of postage that can be paid (below the million dollar limit) is \$999,999.99.

Now let’s assume that the amount paid has to be exact, i.e. we cannot overpay. This is a lot more complicated, as to do this we have to consider all values of q, where q = 17x + 9y + 5z, and where x, y and z are positive integers.  We could write a computer program to check the value of q for all values of x, y and z, and then see which values are missing, and finally pick the highest number among the missing values. We might want to work backwards from \$999,999.99 (since we know we can reach \$999,999.99) and check to see if the a small amount is possible. For example, we know that we can remove one of the 8 stamps of 5 cent denomination, so that gets us to \$999,999.94. We could remove another 5 cent stamp to get to \$999,999.89, and then add a 9 cent stamp to get to \$999,999.98 (x= 5,882,352; y=2; z=6). We can continue this exercise until we hit a number that we cannot create with any combination of values for x, z and z.

The easiest way to solve the problem is to recognize to work backwards to prove that we can calculate most values of q below \$1,000,000:

1. We have defined q=17x+9y+5z
2. We know that q-1=17(x)+9(y+1)+5(z-2)
3. So let’s start with (x=0; y=0, z=20,000,000), whereby q=\$1,000,000
4. Now increase y by 1 and decrease z by 2: q=9,999,999.99
5. Again increase y by 1 and decrease z by 2: q=9,999,999.98
6. Continue this until z reaches zero (x=0; y=10,000,000; z=0), where  q=\$900,000
7. Start again with (x=0;y=0;z=18,000,000), where q=\$900,000
8. Repeat the process of decreasing 1 y by one and decreasing z by 2 (you can do this in an Excel spreadsheet)
9. Continue until z reaches zero, and then reset y to zero and z to whatever is required to reach the last value of q (or a number slightly higher)
10. Repeat the exercise until you reach a value of q = 32 cents.
11. Now you need the 17 cent stamp to get to q=31 (x=1,y=1,z=1).
12. You will also need the 17 cent stamp for q=26 and q=22 (see blow)
13. You can find solutions for q down to 22, but you will get stuck on the number 21…so that must be the answer!

# -2x < 3/7

Monday, September 6th, 2010