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The other side length is equal to



I'll let you do the polynomial division. You should get x - 3 as your result. The units are in feet (since we are looking for a length).

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Yes, we have

C = (5/9)(F - 32)

where F is the temperature in degrees Fahrenheit.

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The expression factors to (n-3)(n^2 - 11n + 31). Since we want a prime number, either n-3 or n^2 - 11n + 31 must be equal to 1 (or -1). Solving for n, we obtain n = 2, 4, 5, or 6. Here are the values we obtain when we replace n into the given expression:

n = 2 --> -13
n = 4 --> 3
n = 5 --> 2
n = 6 --> 3

The values n = 4, 5, 6 produce prime numbers, so the answer is 3.

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If you're trying to solve for x, do whatever operation is necessary to isolate x.

Note the phrase "one-step equation," should only take one step to solve, as opposed to computer algorithms which may take hundreds of steps.

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Given the hint, the problem becomes simple. Since 1 + nt approaches infinity as n approaches infinity, the LHS of the inequality is "bounded" by this expression, so the LHS must also approach infinity.

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No data provided. Also, do you expect tutors on algebra.com to form groups of 2 to 4 people to answer these questions?

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The law of cosines is sufficient. If d is the distance between two ends of the tripod (which is assumed to be increasing), and x is the length of each leg, then



where theta is the angle between two of the legs. It can be shown that if d increases, then theta also increases.

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Points on the boundary line represent the equality cases, or where the two expressions you are comparing are equal. For example, if you know that x + y > 5 (note "strictly greater than 5") and you plug in (3,2), you get 5 > 5, which is not true. However if the inequality were x + y >= 5, (3,2) works.

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No. If f is continuous at x and only x, then for any x-values arbitrarily close to x, f must also be continuous (since continuity is defined using the limit as x approaches a fixed number), so we have a contradiction.

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Assume n > 12 so that the fraction is positive (1 through 11 don't work anyway). If the fraction is reducible for some n, then n-12 and 5n+23 must have a common factor p (other than 1). We can write this using modular arithmetic:

(mod p)

This implies (mod p), so (mod p). Hence, p = 83. Therefore, we can set n = 95, obtaining the fraction 83/498, or 1/6.

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For example, let f(x) = 1 if x is not an integer, but f(x) = 0 if x is an integer (it is easier to write this function using piece-wise notation).

A function f(x) is not continuous if (but not only if)

. This is true because if a is an integer, the LHS equals 1 but the RHS equals 0. Therefore f is not continuous.

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You didn't provide any information whatsoever of the figure. Also, none of the answer choices can be correct because the units for volume are (length)^3.

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Since the product of two of these integers is a power of 2, it follows that both integers are powers of 2. Hence these integers can be {1, 2^2011}, {2, 2^2010}, ..., {2^1005, 2^1006} (order doesn't matter). The two integers cannot be {1, 2^2011} since the third integer would be zero. However, each of the other ordered pairs produces a unique integer (e.g. subtract the sum of the two integers from 2^2011 + 1), so the number of values for the third integer is the number of ordered pairs in {2, 2^2010}, ..., {2^1005, 2^1006}, or 1004 integers.

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B, sometimes. The centroid is defined as the "center of mass" of the triangle, and the circumcenter is not defined this way. However, they can still be the same point, as in the case of an equilateral triangle.

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The only conditions required are that the parallelogram has four congruent sides (or equivalently, equal diagonals) and four right angles.

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Get a 100% and you should be fine. If a 100% is not enough to pass, you're pretty much screwed, that's all I can say.

Seriously though, you shouldn't post all your grades here. If you really want to know what score you need on a final to pass, simply compute the weighted average of your overall grades (final, homework, test, quiz, etc.), letting your final grade be "x." Set this equal to the passing score (presumably 60%) and solve for x. Any score greater than or equal to x will allow you to pass.

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Terms: 5x^8, 3x^6, -2x, 4/5
Coefficients: 5, 3, -2, 4/5
Degree: 8 (highest exponent of x)

I don't know why the problem requires an essay...

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{2,3,4,5,6,7,8,9}

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Suppose I choose 4. The numbers in the pattern are 16, 37, 58, 89, 145, 42, 20, 4, 16, and it repeats.

If the number 1 appears in the sequence, then the previous number must have been 1, 10, 100 or any power of 10. However, all succeeding terms in the sequence must be 1.

Doing the process to the number 37 yields the same sequence as if I did 4 (up to shifting terms).

We can conjecture that every number will eventually produce a periodic sequence. However we want to prove this. Given a sequence a_1, a_2, ... we want to show that a_i = a_j for some distinct i,j (this will produce a periodic sequence). Since it is tricky to establish an explicit formula for a_i, we can instead show that a_i attains a maximum value. Since your sequence can be arbitrarily long, then by Pigeonhole principle, two numbers in the sequence must be the same, and hence the sequence is periodic. I'll leave it to you to prove that a_i attains a maximum value...

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We prove that a convergent sum exists by evaluating the limit as the number of terms approaches infinity:





Since r^(n+1) tends to zero, the sum converges to 1/(1-r). However, this only holds when |r| < 1. Otherwise, the limit diverges.

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Add both equations to get 4y = 12; this implies y = 3. Substitute y with 3 into either equation to obtain x = -4.

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By Vieta's formulas, a+b+c = 2. Hence we are evaluating (2-c)(2-b)(2-a). We could expand this entire expression (since we are guaranteed to obtain useful cyclic sums:





However, we have a+b+c = 2, ab+ac+bc = 3, abc = 4 (you can verify these by expanding (x-a)(x-b)(x-c) and equating like terms). Hence, the expression we want is equal to

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You need to know at least one of the angles (or some side lengths), and use the constraint that the sum of the angles is 180.

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The area is (x-4y)(x+4y) = x^2 - 16y^2 (cm^2).

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Is your sequence kn or k^n? You typed kn, which is interpreted as "k times n" but in question iii) but you implied that the limit as n goes to infinity of kn/k^n (where 0 < k < 1) is zero.

Either way, you can use limits or any other technique. If you mean k^n, part i) is simple, because 1^n is always 1 and 0^n is always 0.

For part ii), we can actually prove it by assuming that on the other hand k^n converges to some number X. If this is so, then k(k^n) must also converge (to kX). However, k(k^n) is equivalent to k^n (since we are evaluating where n approaches infinity) so kX = X. This implies X = 0 (since k is not 1), contradiction. Hence k^n diverges.

For part iii), assume on the other hand it diverges. We have |k(k^n)| < |k^n| (since |k| < 1). This is also a contradiction because we assumed it diverges, so it converges. By the same argument in ii), it converges to 0.

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Several ways to do this. One way is to that

, and that

(modulo 9).

Similarly,

(modulo 9).

Hence the sum is equivalent to 7 + 4 + 1 = 12 = 3 (modulo 9), so the remainder is 3.

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A proof is simply a set of statements that proves why a certain theorem or statement is true.

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A triangle? Or rectangle?

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You'll always get 1 because

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Replace x with 2.5, then simplify to find f(2.5). Do the same with f(12).