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By Pigeonhole principle, if we have ten boxes corresponding to ten pairs of shoes and we have 11 shoes, two shoes must go into the same box (and hence, make a match).

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Assume order of the elements doesn't matter, i.e. {3,4} is equivalent to {4,3}.

Then the number of subsets of size 2 is 7C2 = 7*6/2! = 21.

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Let
f(x) = |x|
g(x) = 6x+9
Then,
f(g(x)) = |6x+9|

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All we know is f(6) = -3 and g(x) = f(3x). The only point on g(x) that we can find occurs when x = 2. This is because g(2) = f(3*2) = f(6) (in other words, I basically set 3x = 6). Since we know f(6) = -3, g(2) must also be -3, so (2,-3) is the corresponding point.

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"I am 4 more than 12" --> 16

That wasn't too hard, was it?

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Do it just like you did with the other numbers. Expected value for this problem is defined as the sum of (profit)*(probability of that profit), so with the case of zero profit, you would add 0*.2, or 0. Since your other numbers are correct, you can just add everything and obtain $500,000.

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g(f(6)) means "g of f(6)," so find f(6).

f(6) = 6^2 + 3 = 39

So g(f(6)) = g(39) = 4*39 = 156

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Volume of a box = length*width*height.

Also, I would think that "approximately 100 - 500 cm3" is an estimate for the volume of a box in cubic centimeters.

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The way algebra.com works is, tutors see a long list of questions (up to 400, 50 by default) posted by users. They will usually answer questions to their liking (e.g. algebra questions, calculus questions, easy/hard questions) or whichever they can answer right away, but it all depends. They have to locate your question and answer it. Since questions are uploaded at a faster rate than tutors can solve it, some questions will go unnoticed.

Best advice would be to repost the question. The site is completely free to use; you can hire certain tutors but that's up to you and them. However, make sure that a tutor can be willing to answer your question. If your question violates the guidelines for posting (e.g. too many questions in one post, too long, ambiguous, poor grammar, etc.), tutors may reply with a negative response, or simply gloss over the question.

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Canceling like factors is similar to canceling common factors in a fraction such as 15/24. Pretty straightforward, as long as you cancel the factors correctly.

Factoring helps solve quadratics, as you don't have to use the quadratic formula. If x^2 + bx + c = 0 factors to (x-p)(x-q) = 0, then right away you know that the roots are p and q. However, not all quadratics factor nicely, and you almost have to "recognize" instantly if a quadratic is factorable (by being able to factor the constant term), otherwise the method is useless.

For example, consider the quadratic . Since 35 = 5*7 and 5+7 = 12, you could factor it like

(check to make sure the coefficients are the same upon expanding). By the zero-product rule, the roots are -5 and -7.

Factoring quadratics with a leading term other than 1 uses a slightly different method; look in your textbook or online for examples.

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For this problem we will assume that .

The best way is probably to try small cases (for m). Let P(m) be the probability that m is the largest number drawn. If m = 1, then every number drawn must be a 1, and



If m = 2, every number must be a 1 or a 2, and we would try to claim something like



However, this is incorrect, because this includes the case where every number is a 1 (which violates the constraint that at least one number must be a 2). To fix this, we subtract P(1):



Finding P(3) is similar to finding P(2), except that we subtract P(2):



This generalizes to

, or



where the plus/minus of (1/n)^k depends on the parity of m (i.e. if m is odd, +; if m is even, -).

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Try to limit your post to one or two questions; here, you posted two questions that have 10 or 12 parts and most tutors won't solve all of them for you. Plus, you didn't provide a quadratic for #1.

For #2, you can approximate it by adding up each f(n), 0 <= n <= d, where f(n) is the function and d is the last day tickets are sold. However, unless you have a graphing calculator that can do summations, this will be quite a task. Another way to approximate it is to take the integral of f(n):



Note the "approximately equal," as the integral assumes f(n) is continuous instead of day-to-day. Only use this method if you know calculus.

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Pythagorean theorem

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"shown below," there's a key phrase. You didn't show any triangle.

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Why don't you post the actual problems instead. Also, you won't learn much from the answers unless you know the solution to them.

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How are we supposed to know where angles a,b,..,e are if you didn't provide a diagram?

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Correlation is defined to be between -1 and 1, so 6 and -6 are impossible. The correlation is 1 if the (x,y) points all lie on a line of positive slope (-1 for negative slope). Since y = 6x - 1 has positive slope, the correlation is roughly 1.

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Just apply the distance formula, which is literally applied Pythagorean theorem.

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I'm pretty sure there's no name for an arbitrary number greater than 1. I would just say it is "a number greater than 1."

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Many different ways. The most common one is
(b = base, h = height)

For the next several formulas, denote

a,b,c = side lengths
A,B,C = angles opposite from sides a,b,c
s = semiperimeter (=(a+b+c)/2)
r = inradius
R = circumradius

Other area formulas you'll see are



(Heron's formula)

where AB and AC are the vectors from A to B and A to C respectively





(excuse my sloppy notation of using A twice...treat them as different variables)

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Reread your question...you stated that the measure of D is 45, and you asked for the measure of D.

You probably want the measure of C, which is 180-45 or 135.

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Look on a unit circle. The sine of an angle is the y-coordinate of the point determined by that angle. The point on the unit circle corresponding to 7 degrees will have some y-coordinate y1, so sin 7 = y1. Similarly, the point corresponding to 353 degrees is "reflected" over the x-axis and will have the y-coordinate -y1, so sin 353 = -y1. y1 + (-y1) = 0, so the statement is true.

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Where are a,b,c,d and angle 1? Also, one angle (angle 1) cannot be supplementary; you need two angles that add up to 180 for them to be supplementary.

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Depends on the distribution of ages in your population.

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Polynomials are defined everywhere, so the domain of P is (-infinity, infinity).

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b is called a subscript.

The log equation is equivalent to



However, (because b^{.01} is greater than b^0 or 1). This implies that cos theta > 1, which cannot occur. Therefore there are no solutions.

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A,B,C can form a triangle so use law of cosines, where a,b,c are the sides opposite A,B,C respectively.



Similarly,





Replace this into your equation and you (should) obtain an identity. However, this part is awfully long so a more elegant solution would be nice. A similar solution would be to use the law of sines:

where R is the circumradius. Therefore

. Also another long solution.

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Because it makes solving them easier, that's the main reason. For example,



Clear fractions by cross-multiplying to get



Same reasoning applies to decimals.

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, where z = 1.645, E = .03, assume p = .5 because this maximizes p(1-p) and we want a safe estimate. Plugging in everything, we obtain



A sample size of 752 would work.

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No, if you rearrange and put in y = mx + b form, you will see that the slopes are the same, and that the graphs of the two lines are parallel.

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For this square matrix, the determinant is equal to

2 det A - 5 det B + 0 det C - 4 det D

where A,B,C,D are the 3x3 matrices whose entries are neither in a row or column of the top entry (e.g. matrix A would contain -2 0 2/-1 1 6/0 3 -2). Additionally, you would have to find the determinant of 3x3 matrices, which reduces to a several 2x2 matrices.

Since this problem requires a lot of "bashing" (not the correct word usage, but math people use it all the time), you could use a calculator. Most graphing calculators can evaluate determinants of square matrices.