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If x=a^p then 1/x=1/(a^p) = a^(-p).
In you case 8=2^3, 1/8 = 1/(2^3) = 2^(-3)

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There are many solutions to this, so I think you are missing something from the problem. The total sum of the page numbers is n(n+1)/2 = 50*101 = 5050, taking away 4949 leaves 101. So the pages that have been ripped out must total 101.
Could do 2 pages with p1 and p100
Could do 3 pages with p11, p40 and p50
Could do 4 pages with p10, p20, p31, p40
...
...
The most number of pages you could have ripped out is 13, that is because the smallest total that can be achieved with 14 pages is the first 14 which totals 105. To find a solution for 13 take the first 13 pages which total 91, so you need another 9, replace page 13 with page 13+9=22.
If you are interested in consecutive pages then notice that any sequence of n consecutive pages will be the first n pages shifted by an integer multiple of n. So you require that n(n+1)/2 + an = 101 has an intrger solution for a. This happens when gcd(n, n(n+1)/2) divides 101. The gcd is n when n is odd, and n/2 when n is even. Because 101 is prime the only divisor is 1, which we get when n=2, therefore the only consecutive page solution is two pages, p50 and p51.
In summary there are lots of solutions, the number of pages ripped out could be any number from 2 to 13. But if you need the pages to be consecutive then is only 2 pages.

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This becomes a standard logic puzzle once you figure out what the number of toys are.
We know that we are looking for 5 numbers which add to 30, all different, all greater than 2. If we take the 4 smallest numbers, then 30 minus that is our largest possible number, 30-(3+4+5+6)=12. So we must choose from the numbers 3-12. We know we need a 5, and we need one number that is 3 times the other. This means we must either take 3,5,9 or 4,5,12. If we take 3,5,9 then we need two more numbers that sum to 13. the only choice is 6,7. If we take 4,5,12 then the remaining two numbers must add to 9, the only choices are 3,6.
So the numbers must be 3,5,6,7,9 or 3,4,5,6,12.
I can't see any way to eliminate either of these choices from the constraints, but every time I try to solve 3,4,5,6,12 it leads to a contradiction. The solution to 3,5,6,7,9 is:
Cher,yellow,train,6
Johnny,green,car,3
Jane ,red,ball,5
Sue,blue,spin,9
Marcia,orange,sled,7
Thanks for this, it was fun. I'd like to see how to eliminate the other set of numbers other than trial and error though. Where did you find it?

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An operation @ is commutative if a@b is the same as b@a. In your problem, using the fact that '+' is commutative, the left hand side can be rewritten as m + 548. This then makes the expression m + 548 = 321 + 548, therefore m must be 321.

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|2x-5| = 6
2x-5 = 6 and -2x+5=6
so 2x=11 => x=5.5 and 2x=-1 => x=-0.5

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Let Q = A or B
P(A or B or C) = P(Q or C) = P(Q) + P(C) - P(Q and C)
P(Q) = P(A) + P(B) - P(A and B)
P(Q and C) = P((A and C) or (B and C)) = P(A and C) + P(B and C) - P(A and B and C)
P(A or B or C) = P(A)+P(B)-P(A and B)+P(C)-(P(A and C) + P(B and C) - P(A and B and C))
P(A or B or C) = P(A)+P(B)+P(C)-P(A and B)-P(A and C)-P(B and C)+P(A and B and C)
P(A or B or C) = 0.39+0.27+0.18-0.13-0.03-0.09+0.01 = 0.6

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If you multiply the equation by 2x then the left hand side has no bottom, if you then multiply the equation by 2x-1 the right hand side will have no bottom. This will give you:
(2x-1)(x+1) = 2x(x-1).
See if you can get anywhere from here.

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How far do you get with this? I've just tried it and I can't spot any tricky or difficult bit, it's just complex algebra. Subtitute z=icot(theta) into the expression (z+1)(z-1), multiply by the complex conjugate of the denominator. Write everything in terms of sin and cos. It should come to cos^2 - sin^2 - 2isincos. Use the double angle formulas to see this is cos2 - isin2, then fiddle about with the minuses (cos -x = cos x, sin -x = -sin x) to get it into the exact form you want.

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To show this we need to label the vertices p, q, and r, such that the vectors pq and pr are the same length and the angle between them is 90 degrees. This statement derives from an imaginary right isosceles triangle with the right angle at p.
If we let D=p, E=q, and F=r, then pq=(3,2) and pr=(2,-3). Both of these vectors have length 13^0.5, and they have a dot product of zero, meaning the angle between them is 90.

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I will assume you mean 8x^2 instead of 8x^3
Well, the first thing to note is that it is a cubic with real coefficients. By the fundamental theorem of algebra it will have 3 zeroes, and because it has real coefficients, then imaginary zeroes have to come in a conjugate pair, so either zero or two of those. The easiest thing to do would be to draw it and see, but let's take a more mathematical approach. We can use the turning points along with the intermediate value theorem to figure out things.
The turning points are the solution to dF/dx = 0. dF/dx = 15x^2+16x-4. Solving this gives approximately 0.21 and -1.28. Now in the function these evaluate to 2.56 and 10.75 respectively.
So lets think about this, the leading term of the function is +x^3 so for sufficiently negative x the function is negative, The function keeps on steadily increasing until it hits the first turning point at -1.28. This has a y value of 10.75 so by the intermediate value theorem, to get from very negative to 10.75 we have a real root, somewhere to the left of -1.28. The function then has to decrease to get down to 2.56 which is the next turning point. Note it is impossible for us to have crossed the axis here because the function is continuous and zero is not between 10.75 and 2.56. After this turning point we increase again and the leading term takes over heading us to +infty. By the IVT there is no zero between 2.56 and +infty so there is only one real zero, and it is negative because it must be less than 1.28.
So in conclusion there is one negative real zero, no positive real zeroes, and two imaginary zeroes. making 3 zeroes in total.

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So the wire is cut into to pieces, some of it is used to make a rectangle, the rest a square. Lets's say the amount used for the rectangle is x. This means the amount used for the square is 12-x.
Lets's consider the square, it has perimeter 12-x, a square has 4 sides, each the same length. So the length of one side of this square will be 3-x/4. So the area of the square will be (3-x/4)^2.
The rectangle has perimeter x, the formula for perimeter is 2*(w+l), but we know from the question that l=w+1. So 2*(w+w+1) = x, 4w+2 = x, w =(x-2)/4. The area of a rectangle is w*l, so in this case it is. (x-2)(x+2)/16.
So the formula for the total area is
(3-x/4)^2 + (x-2)(x+2)/16
I don't like fractions very much, so I am going to simplify this by letting y=x/4
(3-y)^2 + y^2 - 1/4
Looking at this, you can see the dominating term is a positive y^2 (expand the first brackets if you are not convinced). This means the function looks approximately like y^2, which has a minimum at it's turning point. If we were looking for a maximum, then you wouldn't differentiate as the maximum of a y^2 graph is at its endpoints. Of course if it was dominated by negative y^2 the the max would be at the turning point and the min at the end points. Anyway, we have to differentiate to find this turning point.
Remember I'm differentiating wrt x, and dy/dx=1/4. (Use chain rule)
-2*(3-y)/4 + 2y/4 = 0
4y=6
y = 3/2
but remember y=x/4, so x=6. That means the minimum area is when you give exactly half of the string to each.

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v = volume of tank
r = miles per gallon
distance = vr
dr/dv = -0.5 (every increase of 2 decreases by 1)
r = c - 0.5v (integrate)
22 = c - 6 (substitute r=22,v=12)
c = 28
v(28-0.5v) = distance (substitute)
28-0.5v -0.5v = 28-v = 0 (differentiate and solve for turning point)
v = 28
r = 28 - 14 = 14 (calculate r at turning point)
14*28 = 392 (calculate distance)

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P(subscribe) = 0.59
P(dvd) = 0.39
P(subscribe union dvd) = 0.63
P(subscribe union dvd) = P(subscribe) + P(dvd) - P(subscribe intersect dvd)
0.63 = 0.59 + 0.39 - P(subscribe intersect dvd)
P(subscribe intersect dvd) = 0.35

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TP = test positive
TN = test negative
D = diseased
O = not diseased
P(TP | D) = 0.95
P(TN | O) = 0.9
P(D) = 0.01
And obviously (probabilities sum to one)
P(TN | D) = 0.05
P(TP | O) = 0.1
P(O) = 0.99
Want to know:
P(D | TP)
By definition of conditional probability
P(D | TP) = P(D ^ TP)/P(TP)
P(TP | D) = P(TP ^ D)/P(D)
P(D | TP) = P(TP | D)*P(D)/P(TP)
From set theory
P(TP) = P(TP ^ O) + P(TP ^ D)
Conditional probability again
P(TP) = P(TP | O)*P(O) + P(TP | D)*P(D)
Substitute
P(D | TP) = P(TP | D)*P(D)/( P(TP | O)*P(O) + P(TP | D)*P(D) )
0.95*0.01 / ( 0.1*0.99 + 0.95*0.01 ) = 19/217
I'm pretty sure that's right, but last time I did this was 1999!

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We have been told that a system is unstable if their equation has a positive root, so the question is really "do these equations have a positive root."
The second is easier, and we can immediately spot it has a positive root. The leading term is s^3, so for large enough s (denote s_0) it will be positive. You can find such a point if you wish, but we are guaranteed one exists. When s=0 the expression evaluates to -12. The function is continuous so by the intermediate value theorem must have a root between 0 and s_0.
If you are prepared to accept it, then a simple argument for the first problem is to say that the entire expression is made up of multiplication and addition. It is not possible to get 0 by only adding and multiplying positive numbers therefore no positive root.
To do it more rigourously, again, we know there is an s_0 for which the expression is positive, and we know at s=0 the value is 6, so the only way for there to be a positive root, is if there is a positive local minima that is below 0. We can differentiate to find the locations of the extrema. The extrema satisfy

Solving that quadratic (will leave it to you) gives two roots that are both negative, so the extrema are to the left of zero, hence there are no positive roots.

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Ok, there are four questions, each yes or no. Which means each number can be classified into 1 of 16 groups. For example 42 is in the YYYN group (is >25, is div by 2, is div by 3, not div by 5). All of the numbers can be grouped this way.
Now we know that whichever group the number is in has more than one number, because we don't have enough information at this point. We also know that the number isn't a perfect square. So we need a group with more than one number in it, which when the perfect squares are removed only leaves one number.
After you list all the groups, you should find that the group NNNY has 2 members, 5 and 25. When we remove the perfect square (25) that leaves one member, 5, which is therefore the answer. You can also see that there are no other groups that work meaning 5 is the only answer.

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Both printers must have finished 50s or less. The first printer takes 15s per page, so may have printed 0, 1, 2, or 3 pages during this time, which leaves 16, 15, 14, or 13 pages to the other printer. This means the other printer could take any one of 50/16, 50/15, 50/14, or 50/13 seconds per page. I don't think there is enough information to figure out which of those times is correct.

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I am going to use I instead of In for the the identity. So we are given that A^3 - 7A^2 + 5I = 0 therefore A^3 = 7A^2 - 5I.
We are interested in A^4 which is A*A^3, but we know what A^3 is, so substitute A^4 = A(7A^2-5I) = 7A^3-5A. WE now have another A^3 we can substitute for, so do that. A^4 = 7(7A^2-5I)-5A = 49A^2-5A-35I.

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It is neither of those.
Pick some point on the y axis and see what you notice. Can you figure out the equation from that?

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Adding the equations together looks like it may be a good place to start, where exactly are you stuck? Can you show some working?
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I added them and came up with 2x^2-8x+8=0, this is where I was stuck

This is a quadratic equation. If you are given a problem like the one above, you should really be quite comfortable with this type of equation already, and the techniques of how to solve it. You have no idea?
I don't really like the site solver, but try and follow this. However I suggest you go and see your teacher and cover solving quadratics.

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=128 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 4.82842712474619, -0.82842712474619. Here's your graph:

When you have solved for the possible values of x, you must then find y. Substituting for x should leave you some more quadratic equations to solve to give y, you can use the same technique again.

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How far have you got in attempting the problem?
Have you calculated the Lagrangian? Do you know the general form of a Lagrangian dual problem. You need to know that.
Once you know the form of the dual and the Lagrangian the problem should just be trivial algebra, which I'm sure you can do.
You'll have to let me know exactly where you have got stuck before I can give specific help.
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You can wrote the problem so that all 5 constraints are in the form sum(a_i x_i) The Lagrangian for this is g(l,x) = f(x) + sum (l_j sum(a_ij x_i - b_j)) where f is the thing you are trying to maximize and l is >=0.
As you already know we want max_{l_i} min_{x} g(l,x)
g is a linear function of x and therefore is unbounded from below unless all the coefficients of x are zero. When all those coefficients are zero the minimum is what is left over. So we want to max that subject to the conditions that the coefficients of x are zero.
That maximisation is a linear program and is infact the dual program that you are looking for. I think itis normaly written as a minimisation, but to get this simply reaise that maximising the negative of a function is te same as minimizing it.
I hope that wasn't too difficult but I don't know what level to write this at. It really is just as simple as substituting your values into the lagrangian and find the analytical minimum though.

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lets call the new price and the old price . Let's look at the statements one by one
if a price is now 220% more than it was before
Percentages are always over 100, so 220% is 220/100 = 2.2. That's of the old price though, so it is more than the old price. So the new price is 3.2 = 320/100 so the new price is 320% of the old price, yes.
and what it was before is 31.25% of what it is now
What is was before is , and what it is now is . To work out a percentage like this we divide the two things and times by 100. So what we need to do . But we know that from above, we substitute and get There is an on the top and bottom of this division, so it cancels to give 100/3.2 which is 31.25, so again the answer is yes.

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It's not exactly an identity as for x=0 the LHS is not evaluable but the RHS is. Similar for x=pi.
Let S=sin, C=cosine then the first thing I will do is multiply the whole thing by SS(C+1). This gives
-(C+1)(2SC+C-2S-1)=(2S+1)SS
expanding all the brackets gives
-(2SCC + CC - 2SC -C +2SC + C - 2S - 1) = 2SSS + SS
With some cancellations
-(2SCC + CC - 2S - 1) = 2SSS + SS
Now CC + SS =1, so CC = 1-SS, substituting this in gives
-(2S - 2SSS + 1 - SS - 2S - 1) = 2SSS + SS
From here it is just trivial cancellation to see that both are the same.

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Yes, that is indeed wrong. It is easy to check it is wrong because if two expressions are the same they should always equal the same thing, your expression doesn't work for zero, whereas the first one does. I'm not sure how you got that answer, but I suggest you start by factoring the denominator on the left and then see if you can make the denominator on the right look the same.

I.e factor and then see if you can make both denominators look the same.

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Currently you have 2/33 gallons of oil (2 gallons of which 1/33 is oil). You wish to achieve the ratio 40:1 so you need 40 times more gasonline = 40*2/33 = 80/33. Currently you have 64/33 gallons of gasoline in the mixture so you need to add 80/33 - 64/33 = 16/33.

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You agree the volume of this is the cross sectional area times the height, so the real task is to find the cross sectional area.
Draw a circle and on it add a chord of length 10 (the barn wall). Do you agree that the area we want is the major segment defined by this chord?
We know the perpendicular distance from the center to the chord is 5, therefore we can calculate the radius of the circle and the angle defining the relevant sector.
I think you know where to go from here, but if not, you basically say the area of the major segment is the total area minus the area of the minor segment. The area of the minor segment is the area of the sector minus the area of the triangle.

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The circumference of the circle is 60, right? Train A goes at 90mph so after 2.5 hours will have done 225 miles. This is 3 and 3/4 times round the circle, right? So at the starting time of train B, train A is 1/4 or 15 miles behind it.
Train A travels 5 mph faster than B and has to catch up 15 miles, this will take 3 hours.
So at the time they meet, train A will have gone around for 2.5+3=5.5 hours, at 90mph, which is 495miles, or 8 and 1/4 times. Train B goes round for 3 hours at 85mph, a total of 255 miles which is 4 and 1/4 times round.

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So in your working x is the number of adults and y is the number of children? The price of an adult ticket (x) must be two times the price of a child ticket (y) so x=2y. Go on from there and you should get the right answer.

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If we take X gallons of the 25% liquid, and Y gallons of the 35% liquid, then it is quite obvious that we will have X+Y gallons of liquid in total. We need to make 20 gallons of liquid so X+Y=20.
Let us now consider the amount of alcohol in our solution. Taking X gallons of the first liquid will give us 0.25X gallons of alcohol because 25% of the liquid is alcohol. Similarly taking Y gallons of the second liquid gives us 0.35Y gallons of alcohol, so in total there are 0.25X+0.35Y gallons of alcohol in the final mixture.
The final mixture is 20 gallons as we know, so the percentage of alcohol in it is (0.25X+0.35Y)/20. This must be 0.32 (32%) This gives us the second equation (0.25X+0.35Y)/20=0.32. So we now have:
X+Y=20
0.25X+0.35Y=6.4
These equations can be solved for X and Y which should give your answer.

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How have you tried it so far, what did you do to get $45.00?

I can tell you that $45.00 is most certainly wrong. Let's consider a rectangular desktop 2m across and 1.5m deep. The area of this is 2*1.5=3sq m. The blotter for the rectangular table would therefore cost 3*25=$75, and your table is bigger than this rectange, so your answer must be bigger than $75.

Do you know the formula for the area of a trapezium? Once you know this you should be able to work out the area of the desk. Once you know the area of the desk then the price is just the area times $25.

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The initial temperature is when time (x) equals zero. Let's just check I read your equation right, did you mean:



In that case the answer is 60+30=90, yes.