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Sum from n=0 to infinity (5+n)/(6+n)

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Arithmetic means each term changes based on adding or subtracting.
Geometric means each term changes based on multiplication (or division).
1. You are adding 6 to get to each successive term.
2. You are multiplying by 3/2 to get each successive term.
3. You are adding 10 to get to each successive term.
4. You are multiplying by four to get each successive term.
5. You are subtracting 2 to get each successive term.
6. You are multiplying by 3/4 to get each successive term.
Apply the rule I established above and this problem is finished!

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Integrate (find the antiderivative) twice:
f'(x)=2x^2+C
f(x)=(2/3)x^3+Cx+D

Now, f has a horizontal tangent at (2,0) IE the function attains a local extrema and f'(2)=0
hence
f'(2)=2*2^2+C=0
implies C=-8.
Now we have f(x)=(2/3)x^3-8x+D.
But we know f(2)=0, so, (2/3)2^3-8(2)+D=0
16/3-48/3=-D
-32/3=-D
implies 32/3=D.
This gives our final function as
f(x)=(2/3)x^3-8x+32/3.

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For a square pyramid (ie the base is a square...):
Surface Area = [Base Area] +
1/2 × Perimeter × [Side Length]

See http://www.mathsisfun.com/geometry/square-pyramid.html

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First, find where x=x^2 (this gives the end points of the closed interior). This occurs when 1=x and 0=x. So, we are interested in the closed interval [0,1].
Now, you will note by graphing that x>=x^2 throughout [0,1]. IE. .5>.25 etc.
So we integrate from 0 to 1 of (x-x^2)dx.
This is (x^2/2) -(x^3/3) evaluated from 0 to 1 so we have (1/2)-(1/3)=1/6.

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9 consecutive integers:
n,n+1,n+2,n+3,n+4,n+5,n+6,n+7,n+8
So their sum is 9n+36=9(n+4)
So, we want 9(n+4)=9 this implies n+4=1 and n=-3.
So, -3+-2+-1+0+1+2+3+4+5=9.
Now, one should note that 0 is included, so the product of any numbers with 0 has to be 0.

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If by perfect triple is meant Pythagorean triple, we must have the following criteria met (there are more but these should be more than sufficient...):
a^2+b^2=c^2
a=s^2-t^2
b=2st
c=s^2+t^2
with s,t integers. (Note: a and b are interchangeable, IE it doesn't matter which comes first in the first equation.) There are also interesting properties of s and t that one could investigate if interested.
Additionally, |a+b|<=|a|+|b|. IE the hypotenuse is less than or equal to the sum of the legs. This criteria is met by your three numbers, so we must investigate further with the first equation [easiest to do].
The question is: is 6^2+8^2=10^2?
IE is 36+64=100? Yes.
The last two lines are sufficient to understand the concept, but I hope you see there are many more relationships to note with Pythagorean triples than may be presented in a basic algebra or geometry text.

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Any 3 of the same number will work. Also, we can set up something like this
S={2,4,6,}
Then S is multimodal, its median is 4, and its mean is 12/3=4. So we can argue that this set will work, but the multimodality gives something "extra" so it is strictly not perfectly equal.

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Denote a^2=4k+2=2(2k+1). So 2|a^2 thus 2|a. Let a=2n. Then a^2=4n^2 which means 4|a^2.
However, we have a^2=4k+2. 4 does not divide 4k+2, hence 4k+2 can never be a perfect square.

In other words: if a is even, its square will always be divisible by 4.

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Take each side of the equation to the 2/3. 8^(2/3)=4.
So we want x+1=4->x=3. Answer A.

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This is equivalent to log3 (5/9) in your notation. When you subtract logarithms with same base you divide the numbers inside a log.

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You are asking when 2x-5=3 or 2x-5=-3.
IE 2x=8->x=4 or 2x=2->x=1. Two solutions, B.

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Use Euler's phi function. IE phi(n)=n(1-p_1)(1-p_2)...(1-p_r) where each p_i is a constituent prime (powers do not matter).
70=35*2=7*5*2
phi(70)=70(6/7)(4/5)(1/2)=6*4=24.

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Let's denote our odds as:
SUM(k=0->49)[2k+1]
and our evens as:
SUM(k=0->50)[2k]
So, we want to evaluate:
SUM(k=0->50)[2k]-SUM(k=0->49)[2k+1]
this is equivalent to:
2*SUM(k=0->50)[k]-2SUM(k=0->49)[k]-50
which is moreover equivalent to:
2*50+2*SUM(k=0->49)[k]-2SUM(k=0->49)[k]-50
Our summation notation cancels, and we are left with 100-50=50. So we shall pick E as the solution.
If these properties of series are unfamiliar, note first that SUM(k=0->49)[1] is 1 added up 50 times (=50). Moreover, When we easily remove first or last members of the series by simply adding the term, hence 2*SUM(k=0->50)[k]=2*50+SUM(k=0->49)[k].
Hope this helps.

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For polynomial ax^2+bx+c=0. You have a=3, b=5, c=-6. You plug those into the equation above and simplify radicals as necessary.
(1/6)(-5+/-sqrt(97)

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5/3=1.66..., 2pi is "about 2*3.14"=6.28. So, we've got 2,3,4,5,6. 5 numbers D.

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Is the question this:
Is the product of two binomials always a trinomial?
If not, disregard this solution.

Consider:
(a+b)(a-b)=a^2-ab+ab-b^2=a^2-b^2. This is a binomial. Hence, it is not always true that the product of two binomials is a trinomial.

However, it is true in many other cases:
(a+b)^2=a^2+2ab+b^2

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You solve this by substituting x^2 into the expansion of x^5.


--->

D is indeed the answer, but I hope you now see how to do it.

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Consider:
Odd numbers can be expressed as: 2k+1 (k a natural number)
Hence consecutive odds will be 2k+1 and 2k+3.
The sum of these two is 4k+4. Hence, the number must be even so case i is out.
4k+4=400 implies k=99.
So, 400 works. However, as per the tutor below there are additional cases to consider.

Now consider 3 odds.
2k+1,2k+3,2k+5. This gives 6k+9 as the sum. 6k+9=225 -> 6=16. So 225 fits the "or more" portion of the question. I hope this helps you see how to work the problem rather than just supplying the answer...

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The question is this:
Is 2>-2? In words: is 2 bigger than -2? We can add numbers to both sides and retain the inequality. So the question is the same as:
Is 2+2>-2+2? Rather, is 4>0? If you have four of something, don't you have more than if you had none of something?
You can also draw a number line.

-2 -1 0 1 2
The number line by definition adheres to a property called induction which retains the situation that n < n+1

---

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Take out the greatest factors of each term: in this case it is 3x.
3x(x+8)=0
This means 3x=0 or x+8=0 which gives x=0 or x=-8 as solutions.

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Rewrite for y (is easiest):
y < x/2 +3
So, draw the function f(x)=x/2 +3

Now, we want y to always be less than that line (the line itself will be dotted!). So, we shade below it as pictured (fairly poorly but it is better than nothing):

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Is that a log of base 2? By definition log_2(y)=3 means 2^3=y implies 8=y.
In words, the base of a log to the power of the other side of the equation is equal to what the log was of (this may be confusing and it certainly is not concise).

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So, we want:

In other words we are asking when
?
In even more words we are asking when
?
We note that x=0 certainly makes the statement true. Now, will give two real (irrational) roots. So, strictly speaking there are three real numbers that meet the criterion: {0, 1/2 (1 - Sqrt[5]), 1/2 (1 + Sqrt[5]).
However, one should note that x=0 occurs as a root three times.

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There are 21 two digit prime numbers:
11
13
17
19
23
29
31
37
41
43
47
53
59
61
67
71
73
79
83
89
97
Can you make a 9 by adding any two of 1,3,7,9? Hence, one should choose c as the answer.

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First, simplify the left.

So, we want 2=4n-6.
This implies n=2.
If n=2, a=3^(-1)=1/3.
IE

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You can figure this out a couple of ways, but I will show you how I would do it using modular arithmetic and congruences.
First, 7^2=49 which is congruent to (-1) modulo 10.
Second, 7^1992=(7^2)^996 which is congruent to (-1)^996 modulo 10.
Finally, (-1)^996=1. We conclude that the unit digit of 7^1992 is 1.
Without going through an entire explanation and lesson in modular congruences, I will explain this as follows:
Consider:
abcdefx where a,b,c,d,e,f,x are natural numbers that form the number abcdefx (we are not talking about multiplication here). Note that this is the same as abcdef0+x. Now, 10 divides a multiple of 10. So 10|(abcdef0). We are left with x mod 10 (as 10|abcdef0 implies abcdef0 is congruent to 0 mod 10). So when we take mod 10 of any number we can think of that as asking what the last digit will be.
Hope this helps.

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Rewrite as:

Use difference of squares formula.

So we want:
x^2+7x=0 or x^2-7x=0
ie:
x(x+7)=0 or x(x-7)=0
x=-7 or 0; or x=7 or 0

Hence we have solution set {-7,0,7} with 0 occurring twice.

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Please disregard this solution, I misread the question.

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It is asking you to put x=-3 and y=6.
Put them into the inequality and see if it is true.
4(-3)-2(6)=-12-12=-24
Is -24 less than 2? Certainly. Hence, the ordered pair is a solution to the inequality.

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Polynomials have all real numbers as their domains. The question of "what is the domain" is really one asking where is the function defined.
Put 6 into the function. Do you get a "normal" function value? Yes, p(6)=174.

For another example, consider f(x)=1/x. Here we know that the function behaves "funny" around x=0. Moreover, many people are aware of the fact that "we can not divide by zero." Although it takes a bit more math to see why this is true concisely, take a look at the graph and you will see some strange behavior.

A final example, consider g(x)=ln(x). This is the natural log function. ln(x)=y means the same thing as e^y=x. Hence, x is always positive and not 0 as e^y=0 is nonsense.