N-3/8=6
Solution:
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MUx(x,y) = Derivative of U(x,y), treating y constant.
MUx(64,25) =
MUx(49,36) =
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.
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:
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”).
Here’s another more detailed explanation of how to create a proof using mathematical induction:
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.
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:
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: