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�P8�P������{�����C����	�R���C_��@�	kJT�C����	����yi�����C_��@�	k�@�������_��{B�����_�There is some arithmetic to be performed.dummyPerform decimal arithmetic.Calculate the decimal value of a root.Calculate the decimal value of a power.Calculate the decimal value.It may help to factor an integer, for example under a root or square root sign.Evaluate numerically at a point.Compute the decimal value.Compute a numerical value of a function.You can always use approximation methods to find the roots of a polynomial numerically, and hence find its factors, at least to several significant digits.  Choose 'factor polynomial numerically' to have the computer do this.Evaluate a Bernoulli number to a rational numberEvaluate an Euler number to a rational numberConvert a decimal to a fraction.Express a number as a squareExpress a number as a cubeExpress a number as an $n$-th power for a suitable $n$.Express a number as a power of a specified base.Express an integer as a power, for example write $9$ as $3^2$.Express an integer as a sum, using $x = ? + (x-?)$Use the definition of the complex number $i$, namely $i^2 = -1$.Integer powers of the complex number $i$ can be simplified.There is some complex arithmetic to be done.There is a power of a complex number that could be evaluated.Perform complex decimal arithmeticIt may help to factor an integer.Sometimes an integer can be factored into complex factors, like $5 = (2-i)(2+i)$.Factor an expression $n+mi$ into complex factors.  For example, $7-5i = (2-i)(3-i)$.Get rid of the double minus sign.Push the minus sign into the sum.Bring the minus signs out of the sum.When you have a sum containing a sum, you can regroup the terms to remove the extra parentheses.Put the terms in a sum in proper order.You can drop a zero summand using the law $x+0 = x$.There are terms that will cancel out.Collect like terms.Use the commutative law of addition.Pull a minus sign out using $a(b-c) = -a(c-b)$.-ab = a(-b)-abc = ab(-c)a(-b)c = ab(-c)Zero times any number is zero.You can drop a factor of one.Bring out the minus sign using $a(-b) = -ab$Bring out the minus sign using $a(-b-c) = -a(b+c)$Bring out the minus sign using $(-a-b)c = -(a+b)c$Regroup factors to get rid of the extra parentheses, using the associative law of multiplication.When more than one number appears in a product, collect them together at the beginning of the product.Put the factors of a product in standard order.Collect powers, that is, combine terms with the same base into a single term.Multiply out using the distributive law, $a(b+c)=ab+ac$.Use the law for making $(a-b)(a+b)$ into a difference of squares.Expand the square of a sum using a standard formula.Expand the square of a difference using a standard formula.Do you recognize a difference of cubes in its factored form?Do you recognize a sum of cubes in its factored form?Use the commutative law of multiplication.A product of sums, or a power of a sum, can always be multiplied out to get a single sum.  Sometimes this leads to further simplifications, if the original product or power is part of a larger sum.Maybe if you multiply out the numerator, things will get simpler.Maybe if you multiply out the denominator, things will get simpler.Use the operation $na = a + ... + a$.Get rid of the fraction with 0 in the numerator.Get rid of the 1 in the denominator.You have something times its reciprocal here--that makes 1Multiply the fractions to get a single fractionUse the law $a(b/c) = ab/c$ to get a single fractionCancel a common factor from numerator and denominator.Add fractions with the same denominator.Break a fraction with a sum in the numerator apart into two fractions.Break a fraction with a sum in the numerator apart into two fractions, one of which will simplify by cancellation.Use polynomial division to simplify a fraction, when the degree of the numerator is more than the degree of the denominator.You may be able to cancel by polynomial division.Combine numbers in the numerator and denominator into a single rational number using the law au/bv=(a/b)(u/v).Make the denominator into a coefficient using the law $a/b = (1/b) a$Pull out the real factors from numerator and denominator using $au/b = (a/b)u$.Break a fraction apart using $ab/cd = (a/c)(b/d)$.Combine the numerical parts of numerator and denominator into a single coefficient using the law $ab/c = (a/c)b$Cancel the minus signs in numerator and denominator.Push the minus sign into the numerator using the law $-(a/b) = (-a)/b$.Push the minus sign into the denominator using the law $-(a/b) = a/(-b)$.Pull that minus sign out of the numerator so it applies to the fraction as a whole.Pull that minus sign out of the denominator so it applies to the fraction as a whole.Pull the minus signs out of the numerator using the law $(-a-b)/c = -(a+b)/c$.Pull the minus signs out of the denominator using the law $a/(-b-c) = -a/(b+c)$.Adjust the sign of the denominator using the law $a/(b-c) = -a/(c-b)$.Pull the minus signs out of the denominator using the law $-a/(-b-c) = a/(b+c)$.Adjust the sign using the law $-a/(b-c) = a/(c-b)$Pull the minus signs out of the numerator using the law $-(-a-b)/c = (a+b)/c$.Change the order of terms in both numerator and denominator. Select the entire fraction to do this.ab/c = a(b/c)Break a fraction apart using $a/bc = (1/b) (a/c)$.When numerator and denominator are both fractions with the same denominator, you can use the law $(a/c)/(b/c) = a/b$ to get rid of compound fractions.When the denominator is a fraction itself, you invert it and multiply, using the law $a/(b/c)=ac/b$The reciprocal of a fraction is simplified using the law $1/(a/b) = b/a$.When the numerator is a fraction, you can apply the law $(a/b)/c = a/(bc)$ to get rid of the compound fraction.Use $(a/b)/c = (a/b)(1/c)$When the numerator is a product containing a fraction, you can apply the law $(a/b)c/d = ac/bd$Sometimes it helps to factor the denominator.If you have a sum of fractions in the numerator or denominator, you have to first use common denominator to convert that sum to a single fraction.  Then you can proceed to reduce the resulting compound fraction.First factor the denominator, so the true common denominator will be exposed.The denominators aren't the same.  Therefore you must find a common denominator.The denominators aren't the same.  Therefore you must fine a common denominator.  But only add the fractions.You've got a product of fractions, not yet combined into a single fraction.  Multiply your fractions together.You've got something multiplied by a fraction.  Multiply it in, to get a single fraction.It's good housekeeping to keep your factors in the proper order.  It helps to recognize like terms and notice cancellations.Now you have fractions with the same denominator.  They can easily be added to make a single fraction.You have fractions to put over a common denominator.Multiply numerator and denominator by something.You have an exponent of zero.  Get rid of it.You have an exponent of one.  Get rid of it.Zero to any (nonzero) power is zero.One to any power is one.Minus one raised to an integer power can be evaluated: it is 1 for even powers, and -1 for odd powers.You have a power raised to a power.  There is a law for combining such a thing into a single power.You can pull a minus sign out of a power using $(-a)^n = (-1)^na^n$.It might help to push the exponent into the numerator and denominator using $(a/b)^n = a^n/b^n$.You have a power of a product.  It would simplify the expression to push the exponent in using $(ab)^n = a^nb^n$.You can expand the square of a sum using $(a+b)^2 = a^2+2ab+b^2$.The binomial theorem might be fruitful here.You have two or more powers of the same base multiplied together.  Collect those powers.You have a power of a sum; transform it to a product of powers.You have a fraction of the form $a^n/b^n$.  Pull the exponent outside the fraction like this:  $(a/b)^n$.You have powers of the same base in both numerator and denominator.  Combine them into a single power in the numerator.You have powers of the same base in both numerator and denominator.  Combine them into a single power in the denominator.Expand a square.Expand a cube.Expand a power.Break a power into a product of smaller powersExpand a square of a sum.Expand a cube of a sum.Expand a cube of a difference.Use the law $a^(bc) = (a^b)^c$ if $a>0$ or $c\in Z$.Use the law $a^(bc) = (a^c)^b$ if $a>0$ or $c\in Z$.Use the law $a^(bc) = (a^b)^c$, entering the value of $c$.Bring an exponent out of the denominator using $1/a^n = (1/a)^n$Use the definition of a negative exponent, $a^(-n) = 1/a^n$.Negative exponents in the numerator convert to positive exponents in the denominator.Use the definition of a exponent of $-1$, $a^(-1) = 1/a$.Negative exponents in the denominator convert to positive exponents in the numerator.Positive exponents in the denominator convert to negative exponents in the numerator.You can always get rid of a fraction by converting the denominator to a factor with an exponent of -1.A fraction to a negative exponent can be written with a positive exponent after inverting.Use the law $a^(b-c) = a^b/a^c$Combine your product of square roots into a single square root.Make your square root into a product of square roots.You have a squared factor under the square root sign.  Pull it out--but be careful about the sign.The square root of $x^2$ is $x$, at least for positive $x$; but if $x$ is negative, you have to make it the absolute value of $x$.To simplify the square root of an integer, you begin by factoring the integer.The square root of a fraction can be written as a fraction of square roots, using $\sqrt (x/y) = \sqrt x/\sqrt y$The square root of a fraction can be written as a fraction of square roots, using $\sqrt (x/y) = \sqrt |x|/\sqrt |y|$.  The absolute value signs are needed if the signs of $x$ and $y$ are unknown.You have a quotient of square roots. Try to change this into a single square root.Remember that $\sqrt x$ times $\sqrt x$ is $x$.  Therefore  $x/\sqrt x$  simplifies to $\sqrt x$.Remember that $\sqrt x$ times $\sqrt x$ is $x$.  Therefore  $\sqrt x/x$  simplifies to $/\sqrt x$.An even power of a square root can be simplified using $(\sqrt x)^2^n = x^n$, at least for nonnegative $x$An odd power of a square root can be simplified using $(\sqrt x)^(2n+1) = x^n\sqrt x$.Perhaps the square root can be evaluated exactly?Evaluate the square root using decimal numbersDo the numerator and denominator have a common factor under the square root sign?Factor the polynomial under the square root sign.Rationalize the denominator.  That means to multiply numerator and denominator by the same thing, chosen in order to get rid of square roots in the denominator.Rationalize the numerator.  That means to multiply numerator and denominator by the same thing, chosen in order to get rid of square roots in the numerator.A square root of an even power can be simplified using absolute valueThere is a common factor under the square roots in numerator and denominator.  Cancel the common square root.Multiply out under the square root sign.It may help to think of $b$ as the square of $\sqrt b$, so $a^2-b = (a-\sqrt b)(a+\sqrt b)$.A root with index 2 should be converted to a square root.Express a square root as a root of a power, for example $\sqrt 2 = ^4\sqrt 4$Express a square root as a power of a root, for example $\sqrt 3 = (^4\sqrt 3)^2$An even power is a square, so you have a square under the square root sign.You have a power more than two under the square root sign; bring some powers outside the square root.Push something under the square root using $a\sqrt b = \sqrt (a^2b)$.Rationalize the denominator and simplify.An exponent of $\onehalf $ can be converted to a square root.A fraction in the exponent with denominator 2 can be converted to a square root, using  $a^(n/2) = \sqrt (a^n)$.A fraction in the exponent with denominator $n$ can be converted to an $n$-th root, using $a^(b/n) = ^n\sqrt (a^b)$.A square root can be converted to an exponent of $\onehalf $An $n$-th root can be converted to an exponent of $1/n$Eliminate roots of powers by changing to fractional exponents.Eliminate powers of roots by changing to fractional exponents.Eliminate powers of square roots by changing to fractional exponents.An $n$-th root in the denominator can be converted to a negative exponent of $1/n$Express a square root in the denominator using a negative fractional exponent.Powers of $-1$ can be explicitly evaluatedFactor an integer which is raised to a fractional exponentGet the fractional exponent out of the denominator.Get the fractional exponent out of the numerator.Make the fractional exponent into a power of a square rootMake the fractional exponent into a power of a rootCombine the product of roots into a single root.Split the root of a product into a product of roots.Bring the exponent outside the root so everything is a function of the same root.You have an $n$-th power under an $n$-th root. Pull it out.An $n$-th root of an $n$-th power can be simplified, but be careful: $^n\sqrt (x^n) = x$ isn't always true!You can simplify the root: for example the cube root of $x^6$ is $x^2$Sometimes you can lower the index of a root.  For example, the 6-th root of $x^3$ is $\sqrt x$.Sometimes you can lower the index of a root.  For example, the 6-th root of $x^2$ is the cube root of $x$.Remember the definition of the $n$-th root of $x$:  when you raise it to the $n$-th power, you get $x$.You have a power of a root.  Bring the exponent under the root, as in $(^n\sqrt x)^2 = ^n\sqrt (x^2)$.You have a power of a $n$-th root, say of $x$.  Pull out some factors of $x^n$ until the power is less than $n$.  Example:  $(^3\sqrt 2)^7 = 2^2 ^3\sqrt 2$.Factor the integer under the root sign.You have an odd root of a negative expression; bring the minus sign out from under the root.Maybe the root can be evaluated exactly.Factor the polynomial under the root sign.Multiply out under the root sign.A square root of a square root can be expressed as a fourth root.A square root of an n-th root can be expressed as a 2n-th root.An n-th root of a square root can be expressed as a 2n-th root.A root of a root can be expressed as a single root.  For example, a cube root of a fourth root is a 12-th root.Turn your root of a fraction into a fraction of roots.Turn a quotient of two roots into a single root.Combine the roots in numerator and denominator, getting a single root.Cancel a factor under the root sign.  Select the whole fraction.Cancel a root from numerator and denominator.  Select the whole fraction.The numerator and denominator have a common factor under the root sign.  Select the whole fraction.Push something under the root using $a\sqrt b = \sqrt (a^2b)$.Push a minus sign under a root.Bring the whole fraction under the root.Bring the whole fraction under the square root.A power of a root can be simplified, making it a root with a lower indexA power of a root can be simplified, creating a square root.You know that $i^2$ is $-1$.  It follows that $1/i$  is $-i$.Since $1/i$ is $-i$,  $i$ can be brought up from the denominator to the numerator if you change the sign of the fraction.By definition, the square root of $-1$ can be written as $i$.The square root of a negative number can be expressed in terms of $i$, using the law $\sqrt (-a) = i\sqrt a$.You can clear $i$ out of the the denominator entirely, by multiplying numerator and denominator by the complex conjugate of the denominator.A complex number times its conjugate simplifies according to $(a-bi)(a+bi) = a^2+b^2$.You can factor a sum of squares using complex numbers, according to $a^2+b^2 = (a-bi)(a+bi)$.By the Pythagorean theorem, we have $|u + vi|^2 = u^2 + v^2$By the Pythagorean theorem, we have $|u + vi| = \sqrt (u^2+v^2)$Express the quotient as a single complex number, using $(u+vi)/w = u/w + (v/w)i$.Write complex numbers in the form $u+vi$Express a complex square root to the form $u+vi$Factor out a number.Clear your numerical denominators, so you can see better what is going on.There is a common factor that you could pull out using the distributive law, $ab+ac = a(b+c)$Factor out the highest common power.Do you see the perfect square of a sum? Remember $a^2+2ab+b^2 = (a+b)^2$.Do you see the perfect square of a difference? Remember $a^2-2ab+b^2 = (a-b)^2$.A difference of squares can be factored using $a^2-b^2 = (a-b)(a+b)$.This doesn't seem to fit any of the simpler patterns, but it is a quadratic trinomial, so maybe it can be factored.If it won't factor any other way, you could always use the quadratic formula on it.An even power can be written as a square, using $a^2^n = (a^n)^2$.  Then maybe you can use factoring patterns involving squares.Try combining powers using the law $a^nb^n = (ab)^n$It may help to factor the coefficients of your polynomial.Factor that integer.It may help to make a substitution.Now eliminate your defined variable.Regard a variable as constant.This is too complicated to factor directly, but if you write it as a function of some subexpression, correctly, you may make progress.Express a higher power as a cube using the formula $a^(3n) = (a^n)^3$Express a power using the formula $a^(mn) = (a^m)^n$.There is a formula for factoring the difference of cubes.There is a formula for factoring the sum of cubes.There is a formula for factoring $a^n-b^n$.There is a formula for factoring $a^n+b^n$.There are formulae for factoring the sum of fourth powers.Some fourth-degree polynomials can be factored by special formulas.Try making a substitution.  Select the term to be replaced by a new variable.Guess a factor.If all else fails, you can search for a linear factor systematicallyTry to factor by groupingWrite it as a polynomial in some variable or expression.  Select the variable or expression.Switch sides, in order to get the unknown on the left.Change the signs of both sides.Add something to both sides of your equation.Subtract something from both sides of your equation.Transfer an appropriate term from left to right.Transfer an appropriate term from right to left.Multiply both sides of your equation by something.Divide both sides of your equation by something.Square both sides of your equation.Cancel a term from both sides of your equation.Cancel a common factor of the two sides of your equation.Subtract to put in form $u=0$.When an equation reduces to an identity, any number (for which the sides are defined) is a solution.  The equation reduces to the logical expression 'true'.When the two sides of an equation have opposite signs, the only way the equation can have a solution at all is if both sides are zero.  That is, $a = -b$ becomes $a^2 = -b^2$, provided $a$ and $b$ are both non-negative.  This way of writing the equation will often save you from generating false solutions that have to be rejected in the end.When the two sides of an equation have opposite signs, the only way the equation can have a solution at all is if both sides are zero.  That is, $a = -b$ becomes $a=0$, provided $a$ and $b$ are both non-negative.  At the end, you check the solution, and if $b$ was not also zero, the solution will be rejected.When the two sides of an equation have opposite signs, the only way the equation can have a solution at all is if both sides are zero.  That is, $a = -b$ becomes $b=0$, provided $a$ and $b$ are both non-negative.  At the end, you check the solution, and if $a$ was not also zero, the solution will be rejected.You have a product equal to zero.  Split that into two (or more) equations setting each factor to zero, using the law:  if ab=0 then a=0 or b=0.You can always use the quadratic formula on any quadratic equation.Complete the square.Take the square root of both sides.You have an equation of fractions, with no obvious simplifications, so the thing to do is cross multiply.If the discriminant is negative, a quadratic has no real roots.You have two equations of the form $u^2 = a$ and $u^2 = -a$.  They can be combined into $u^2 = |a|$.You could choose 'solve numerically' to let the computer find solutions using an iterative approximation method.You could raise both sides to a power, using the law, if $u=v$ then $u^n=v^n$.In order to get at the unknown under the square root, take the square root of both sides.In order to get at the unknown under the root, take the $n$-th root of both sides.In order to get at the unknown, apply a suitable function to both sides.Put your fractions over a common denominator.Split your equation into two or more equations using the law: if ab=0 then a=0 or b=0Split your equation into two or more equations using the law: if ab=ac then a=0 or b=cSelect one equation.Show all your equations again, you're finished with the one that is visible.Collect multiple solutions.Maybe you could make a helpful substitution.  Select the expression to be replaced by a new variable.One of your equations is impossible--reject it.Don't forget to check the roots in the original equation.You could solve that linear equation at once.Make an appropriate substitution to eliminate the quadratic term.The discriminant determines whether there are 3 real roots or only 1, and you have to compute it first to know which cubic formula to apply.You must show the cubic equation again in order to continue working on it.As Vieta discovered in 1592, when solving $cx^3 + ax + b = 0$, you can substitute $x = y - a/(3cy)$, which will produce an equation quadratic in $y^3$. Select the whole equation to see that choice.Your cubic has only one real root, because the discriminant is positive.Your cubic has three real roots, because the discriminant is negative.Make a substitution $x = f(u)$ where $x$ is an old variable and $u$ is new.Now it is time to get rid of the defined variable.These two expressions will be the same if you change one of the integer variables.  Select one of the integer variables and make a substitution.  After that one equation will drop out.  At present each equation stands for three roots, so there are apparently six roots, but really there are only three.Evaluate the expressions for the roots to get the exact answers.The best you can do is find approximate decimal values for the rootsSimplifyTry to get the logarithm in the exponent using the law:  if $u=v$ then $a^u = a^v$.Get rid of the logarithm on the left side using: if $ln u = v$ then $u = e^v$.Get rid of the logarithm on the left side using: if $log u = v$ then $u = 10^v$.Get rid of the logarithm on the left side using: if $log(b,u) = v$ then $u = b^v$.Since both sides are powers, and the bases are the same, the exponents must be equal too.Take the log of both sides.Take the ln of both sides.One of your equations is impossible--remember logarithms of negative numbers aren't defined.Use Cramer's ruleEvaluate the determinant.  MathXpert will do that for you in one easy step.First get the variables on the left side and the constants on the right side.Collect like terms, so that you have only one term in each variable.Line up the variables nicely, so you can compare the coefficients in different equations easily.Add two equations.Subtract two equations.Multiply an equation by a constant.Divide an equation by a constant.Add a multiple of an equation to another equation.Subtract a multiple of an equation from another equation.Swap two equations.Put the solved equations in order.Drop an identity.Regard a variable as constant, so as to solve for the rest.Can these equations actually be solved?  It seems you may have a contradiction on your hands.Add two equationsSubtract two equationsSolve one of the unsolved equations for one variable in terms of the rest.Add two rows.Subtract one row from another.Multiply some row by a constant.Divide some row by a constant.Add a multiple of one row to another row.Subtract a multiple of one row from another row.Swap two rows.Write a matrix $A$ as the product $IA$, where $I$ is the matrix identity.  Then when you perform row operations, the inverse of $A$ will develop where $I$ is.Simplify one or more of your equations.Cancel a term that occurs on both sides of one of your equations.Add something to both sides of one of your equations.Subtract something from both sides of one of your equations.Divide one of your equations by a constant to isolate a variable.After you have expressed one variable in terms of the rest, use that equation to eliminate that variable from the other equations, by substituting for that variable.Your equations are contradictory.To begin with, write your equations in matrix form.Multiply the right side by the matrix identity $I$.Multiply matrices.A column entirely consisting of zeroes can be dropped.A row consisting entirely of zeroes can be dropped.A duplicate row can be dropped.A matrix equation can be converted to a system of ordinary equations.Solve using a symbol for the matrix inverse:  $AX = B  =>  X = A^(-1)B$There is an explicit formula for a 2 by 2 matrix inverse.Ask MathXpert to compute the exact matrix inverse.  Select the matrix inverse you want to compute.You could ask MathXpert to compute the decimal matrix inverse.  Select the matrix inverse you want to compute.For nonnegative $u$, you can get rid of absolute value signs using $|u| = u$.You could always assume $u\ge 0$ and set $|u| = u$.For negative $u$, you can get rid of absolute value signs using $|u| = -u$.You can pull a nonnegative quantity out of absolute value using $|cu| = c|u|$.You can get a positive denominator out of absolute value using $|u/c| = |u|/c$.You can simplify a product of absolute values using $|u||v| = |uv|$.If it helps, you can break up an absolute value using $|uv| = |u||v|$.Push absolute values into numerator and denominator using $|u/v| = |u| / |v|$.Get absolute values out of your fraction using $|u| / |v| = |u/v|$Even powers of absolute value can be simplified using $|u|^2^n=u^2^n$ if $u$ is real.Absolute values of a power obey the law $|u^n|=|u|^n$ if $n$ is real.Absolute values of square roots obey the law $|\sqrt u| = \sqrt |u|$.Absolute values of roots obey the law $|^n\sqrt u| = ^n\sqrt |u|$.You can cancel under the absolute value signs using the law $|ab|/|ac|=|b|/|c|$You can cancel under the absolute value signs using the law $|ab|/|a|=|b|$Maybe there is a common factor in what is inside the absolute values in numerator and denominator.  If so it would be helpful to show it explicitly.If $c\ge 0$, you can always split an equation $|u|=c$ into the two equations $u=c$, $u = -c$.The equation $|u|/u=c$ has real solutions $u$ only when $c$ is 1 or $-1$, and then the solutions are $u = 1$, $u = -1$.For $v\ge 0$, $|u| < v$ iff $u$ is (strictly) between  $-v$ and $v$For $v\ge 0$, $|u| \le  v$ iff  $u$ is between  $-v$ and $v$$u < |v|$ iff $v < -u$ or $u < v$$u \le  |v|$ iff $v \le  -u$ or $u \le  v$An equation $|u| = u$ can be converted to an inequality $0 \le  u$, eliminating the absolute value sign.An equation $|u| = -u$ can be converted to an inequality $u \le  0$, eliminating the absolute value sign.An absolute value can't be negative: $0 \le  |u|$ is always true.An absolute value can't be negative: $|u| < 0$  is always false.An absolute value can't be negative: $-c \le  |u|$ is always true provided $c$ is nonnegative.An absolute value can't be negative: $-c < |u|$ is always true provided $c$ is positive.An absolute value can't be negative: $|u| < -c$ is false, provided $c$ is nonnegativeAn absolute value can't be negative: $|u| \le  -c$ is false, provided $c$ is positiveIf $c \ge  0$, the inequality $|u| \le  -c$ is possible only if $u$ and $c$ are both zero.  In MathXpert, you handle this situation by using $|u| \le  -c$ iff $u=0$ assuming $c=0$.  The assumption $c=0$ will be made.  If it eventually contradicts $u=0$ there will be no solution.  Otherwise you will find the solution by solving $u=0$.If $c \ge  0$, the equation $|u| = -c$ is possible only if $u$ and $c$ are both zero.  In MathXpert, you handle this situation by using $|u| = -c$ iff $u=0$ assuming $c=0$.  The assumption $c=0$ will be made.  If it eventually contradicts $u=0$ there will be no solution.  Otherwise you will find the solution by solving $u=0$.$v>|u|$ iff $u$ is (strictly) between  $-v$ and $v$$v\ge |u|$ iff  $u$ is between  $-v$ and $v$$|v|>u$ iff $-u>v$ or $v>u$$|v|\ge u$ iff $-u\ge v$ or $v\ge u$Absolute values are always nonnegative.An absolute value cannot be negative.If $c \ge  0$, the inequality $-c \ge  |u|$ is possible only if $u$ and $c$ are both zero.  In MathXpert, you handle this situation by using $|u| \le  -c$ iff $u=0$ assuming $c=0$.  The assumption $c=0$ will be made.  If it eventually contradicts $u=0$ there will be no solution.  Otherwise you will find the solution by solving $u=0$.You can write an even power as a power of an absolute valueAbsolute values of a power obey the law $|u|^n = |u^n|$ if $n$ is real.$u < v$ means the same as $v > u$Add a suitable term to both sides of your inequality.Subtract a suitable term from both sides of your inequality.Change the signs of both sides, but remember that will change the direction of the inequality:  -u < -v =>  v < uYou can change the signs of both sides, but you must change $<$ to $>$ at the same time.You can multiply both sides of an inequality by the same quantity $c$.  But the sign of $c$ must be known, and if you only know $0 \le  c$ you will give up $<$ for $\le $.If you would like to multiply both sides by something, but you don't know whether it is positive or negative, you can always multiply by its square instead, since that is always non-negative.You can divide both sides of an inequality by the same quantity $c$.  But the sign of $c$ must be known.When both sides are numbers, you can just evaluate the inequality numerically.A square, or any even power, is always non-negative.A square, or any even power, can never be negative.Square both sides, which is legal since both sides are nonnegative.Square both sides.  Since the smaller side isn't obviously non-negative, you will get an extra inequality to account for the possibility that it is negative.You have an inequality $u < v$ and the corresponding equation $u = v$; combine them.Two of your solutions define overlapping intervals.  Combine those intervals.You have one or more solutions that do not satisfy the original inequality.  Such solutions can be introduced by squaring an inequality or cancelling an expression.  Use the assumptions to reject or correct these solutions.$u > v$ means the same as $v < u$You can change the signs of both sides, but you must change $>$ to $<$ at the same time.You can change the signs of both sides and keep the same inequality sign by changing $-u > -v$ to $v > u$.A square, or any even power, is always non-negativeYou have an inequality $u > v$ and the corresponding equation $u = v$; combine them.$x \le  y$ means the same as $y \ge  x$Change the signs of both sides, but remember that will change the direction of the inequality.You can change the signs of both sides and keep the same inequality sign by changing $-u \le  -v$ to $v \ge  u$.You can multiply both sides of an inequality by the same quantity, but you must know the sign, because $\le $ must change to $\ge $ when you multiply by a negative quantity.You can divide both sides of an inequality by the same quantity, but you must know the sign, because $<$ must change to $>$ when you divide by a negative quantity.$x \ge  y$ means the same as $y \le  x$You can change the signs of both sides, but you must change $\ge $ to $\le $ at the same time.You can change the signs of both sides and the same inequality sign by changing $-u \ge  -v$ to $v \ge  u$.You can take the square root of both sides, but you have to be careful: $u^2 < a => |u| < \sqrt a$.  Don't forget the absolute value.Take the square root of both sides;  you should get an interval between the two square roots of the constant side.You can take the square root of both sides, but you have to be careful: $0 \le  u < v^2 => \sqrt u < |v|$When you take the square root of this inequality, you will get two inequalities, corresponding to the positive and negative square roots.Squares are always non-negative, so the first inequality can be dropped. Select the whole inequality to do this.Get rid of a square root or absolute value by squaring both sides of your inequality.You can take the square root of both sides of an inequality if you know everything is nonnegative:  $0 \le  u < v => \sqrt u < \sqrt v$Squares are always non-negative.Square roots are always non-negative, but if you square a square root, don't forget that what is under the root must be nonnegative.You can take the square root of both sides, but you have to be careful:  $u^2 < a => |u| < \sqrt a$.  Don't forget the absolute value.You have a square root.  Get rid of it by squaring both sides of your inequalityTake the reciprocal of both sidesTake the reciprocal to get the unknown out of the denominator.Take the reciprocal, but be careful when the interval includes zero!You can take odd roots of both sides of any inequality.You can take even roots of both sides, but you have to be careful:  $u^2^n < a => |u| < ^2^n\sqrt a$.You can take even roots of both sides, but you will get a part corresponding to the negative root: $u^2^n < a$ iff $-^2^n\sqrt a < u < ^2^n\sqrt a$.You can take even roots of both sides, but you have to be careful: $0 \le  a < u^2^n => ^2^n\sqrt a < |u|$.You can teke even roots of both sides, but you will get a part corresponding to the negative root:  $a < u^2^n$ iff $v < -^2^n\sqrt a$  or $^2^n\sqrt a < u$.You can take an even root of all three terms, but you will get an extra interval corresponding to the negative roots.You have an $n$-th root.  Get rid of it by raising both sides to the $n$-th power. But remember that even roots of negative numbers are not defined, so you must explicitly keep that condition.  For example, $^4\sqrt x < 16$ becomes $0 \le  x < 2$.You have an $n$-th root.  Get rid of it by raising both sides to the $n$-th power.You can always raise both sides of any inequality to a positive odd power.You can raise both sides of an inequality to any positive power, if both sides are known to be nonnegative.Even-index roots are always non-negative, but if you raise such a root to a power, don't forget that what is under the root must be nonnegative.You can take even roots of both sides, but you have to be careful:  $u^2^n \le  a$ iff $|u| < ^2^n\sqrt a$.You can take even roots of both sides, but you will get a part corresponding to the negative root: $u^2^n \le  a$ iff $-^2^n\sqrt a \le  u \le  ^2^n\sqrt a$You can take even roots of both sides, but you have to be careful: $0 \le  a \le  u^2^n$ iff $^2^n\sqrt a \le  |u|$You can teke even roots of both sides, but you will get a part corresponding to the negative root:  $a \le  u^2^n$ iff $v \le  -^2^n\sqrt a$  or $^2^n\sqrt a \le  u$.You have an $n$-th root.  Get rid of it by raising both sides to the $n$-th power. But remember that even roots of negative numbers are not defined, so you must explicitly keep that condition.  For example, $^4\sqrt x \le  16$ becomes $0 \le  x \le  2$.You should drop any positive factors.The numerator is positive, so the fraction is positive if and only if the denominator is positive.In $0 < u/\sqrt v$, multiply by $v\sqrt v$, not just $\sqrt v$, or you will lose domain information.  Note that $v\sqrt v$ is positive. The square roots will cancel out.$u/v$ is positive if and only if $u$ and $v$ have the same sign.  That's the same condition for $uv$ to be positive, and $0 < uv$ may be easier to work with than $0 < u/v$.In $u/\sqrt v < 0$, multiply by $v\sqrt v$, not just $\sqrt v$, or you will lose domain information.  Note that $v\sqrt v$ is positive. The square roots will cancel out.$u/v$ is negative if and only if $u$ and $v$ have opposite signs.  That's the same condition for $uv$ to be negative, and $uv < 0$ may be easier to work with than $u/v < 0$.In solving a linear inequality, it may help to factor out the coefficient of the unknown:  $ax \pm  b < 0$ iff $a(x\pm b/a) < 0$.When you have an inequality of the form $(x-a)(x-b) < 0$, the solution set is the interval between the zeroes of the quadratic, that is, $a < x < b$, if $a < b$.When you have an inequality of the form $0 < (x-a)(x-b)$, say with $a < b$, the solution set is composed of all values not between the two roots, that is,  $x < a$ or $b < x$.The numerator is positive, so the fraction is non-negative if and only if the denominator is non-negative.In $0 \le  u/\sqrt v$, multiply by $v\sqrt v$, not just $\sqrt v$, or you will lose domain information.  Note that $v\sqrt v$ is positive. The square roots will cancel out.$u/v$ is positive if and only if $u$ and $v$ have the same sign.  That's the same condition for $uv$ to be positive, and $0 \le  uv$ may be easier to work with than $0 \le  u/v$.In $u/\sqrt v \le  0$, multiply by $v\sqrt v$, not just $\sqrt v$, or you will lose domain information.  Note that $v\sqrt v$ is positive. The square roots will cancel out.$u/v$ is negative if and only if $u$ and $v$ have opposite signs.  That's the same condition for $uv$ to be negative, and $uv \le  0$ may be easier to work with than $u/v \le  0$.$u \le  v => v \ge  u$When you have an inequality of the form $(x-a)(x-b) \le  0$, the solution set is the interval between the zeroes of the quadratic, that is, $a \le  x \le  b$, if $a < b$.When you have an inequality of the form $0 \le  (x-a)(x-b)$, say with $a < b$, the solution set is composed of all values not between the two roots, that is,  $x \le  a$ or $b \le  x$.You can take the square root of both sides, but you have to be careful:  $a > u^2$ becomes $\sqrt a > |u|$.  Don't forget the absolute value.You can take the square root of both sides, but you have to be careful: $v^2 > a$ becomes $|v| > \sqrt a$ provided $a > 0$.You can take the square root of both sides of an inequality if you know everything is nonnegative:  $0 \le  u < v$ implies $\sqrt u < \sqrt v$You can take the square root of both sides, but you have to be careful:  $a \ge  u^2$ becomes $\sqrt a \ge  |u|$.  Don't forget the absolute value.You can take the square root of both sides, but you have to be careful: $0 \le  u < v^2$ becomes $\sqrt u < |v|$You can take even roots of both sides, but you have to be careful:  $a > u^2^n$ becomes $ ^2^n\sqrt a > |u|$.You can take even roots of both sides, but you will get a part corresponding to the negative root: $ a > u^2^n$ iff $-^2^n\sqrt a < u < ^2^n\sqrt a$.You can take even roots of both sides, but you have to be careful: $0 \le  a < u^2^n$ becomes $^2^n\sqrt a < |u|$.You can take even roots of both sides, but you have to be careful:  $u^2^n \le  a iff |u| < ^2^n\sqrt a$.You can take even roots of both sides, but you will get a part corresponding to the negative root: $u^2^n \le  a$ iff $-^2^n\sqrt a \le  u \le  ^2^n\sqrt a$.You can take even roots of both sides, but you have to be careful: $0 \le  a \le  u^2^n $iff $^2^n\sqrt a \le  |u|$.You can teke even roots of both sides, but you will get a part corresponding to the negative root:  $a \le  u^2^n$ iff $ v \le  -^2^n\sqrt a$  or $^2^n\sqrt a \le  u$.In $u/\sqrt v \le  0$, multiply by $v\sqrt v$, not just $\sqrt v$, or you will lose domain information.  Note that $v\sqrt v $is positive. The square roots will cancel out.Expand the power, using the binomial theorem.Use the binomial theorem in the form with the binomial coefficients $(n k)$.Express the binomial coefficients in terms of factorials, using $(n k) = n!/((n-k)!k!)$.Use the definition of factorial, $n! = n(n-1)(n-2)...1$.Compute the factorials explicitly.Evaluate the binomial coefficients (n k).Expand the $\sum $ notation to an ordinary sum.Evaluate the sum written in $\sum $ notation to a rational number.Use the recursion equation for the factorial function, $n! = n(n-1)$.$n!$ is divisible by $n$, with quotient $(n-1)!$.$n!$ is divisible by $(n-1)!$, with quotient $n$.$n!$ is divisible by $k!$ when $k$ is less than $n$.Do you recognize the cube of a sum?  Factor it.Do you recognize the cube of a difference?  Factor it.Do you recognize the fourth power of a sum?  Factor it.Do you recognize the fourth power of a difference?  Factor it.Do you recognize a power of a sum?  Factor it.Do you recognize a power of a difference?  Factor it.The summand doesn't depend on the index variable, so the sum is just the summand times the number of terms.Try to get the minus sign outside the $\sum $ sign.Pull constants outside the $\sum $ signBreak the sum into two or more sums using $\sum (u+v) = \sum u + \sum v$Break the sum into two sums using $\sum (u-v) = \sum u - \sum v$Expand the sum written using $\sum $  as an ordinary sum, written with $+$.There is a formula for the sum of the first $n$ integers.There is a formula for the sum of the first $n$ squares.There is an explicit formula for the sum $1+x+..+x^n$.Show the first few terms.Evaluate the sum written in $\sum $ notation to a rational.Evaluate to decimal.Express the summand as a polynomial in the index variable.This is a telescoping sum:  part of each term cancels with the next term.Shift the summation index; that is, add something to both lower and upper limits and change the sum accordingly so it still represents the sum of the same terms.Rename the index variable.A product of two sums converts to a double sum:  $(\sum u)(\sum v) = \sum  \sum  uv$Split off the last term of the sum, so as to be able to use the induction hypothesis.There is a formula for the sum of the first $n$ cubes.There is a formula for the sum of the first $n$ fourth powers.You can differentiate term by term.  That is, the derivative of a sum is the sum of the derivatives.Pull the derivative out of the sum.  Select the whole sum to see this choice.You can integrate term by term.  The integral of an indexed sum is the sum of the integrals.Pull the integral out of the sum.  Select the whole sum to see this choice.Push a constant into a sum.If the lower index of summation were zero, you would be able to solve this.If the lower index of summation were different, you would be able to solve this.Select the induction variable.Start with the basis case.Start your induction step.Now use your induction hypothesis.You've got all the pieces.  Just draw your final conclusion!Remember that the sin function takes values between $-1$ and 1:  $|sin u| \le  1$Remember that the cos function takes values between $-1$ and 1:  $|cos u| \le  1$$sin u \le  u$  if $u\ge 0$$1 - u^2/2 \le  cos u$By definition of the arctan function, we have $|arctan u| \le  \pi /2$$arctan u \le  u$ if $u\ge 0$$u \le  tan u$  if $u\ge 0$You can take the ln of any inequality (if the sides are positive).You can take the log of any inequality (if the sides are positive).Try to eliminate logarithms by taking powers.exponentials dominate polynomialsalgebraic functions dominate logarithmsRemember that log $a$ is the number such that $10^log a = a$.A log in the exponent can be simplified using the law: $10^(n log a) = a^n$Remember $log 10^n = n$, at least for $n$ real.Remember the logarithm of 1 is 0.Remember log 10 is 1.Express log in terms of ln using the conversion formula: $log a = (ln a)/(ln 10)$.Any power $u^v$ can expressed using logarithms as  $10^(v log u)$If you factor a number, you can then break up its logarithm.You can simplify a logarithm by factoring out powers of 10.log(a/b) = -log(b/a)log(b,a/c) = -log(b,c/a)Break logs of powers up using $log a^n = n log a$.To multiply, add logarithms:  $log ab = log a + log b$The log of the reciprocal is the negative log:  $log 1/a = -log a$To divide, subtract logarithms:  $log a/b = log a - log b$To multiply, add logarithms:  $log a + log b = log ab$To divide, subtract logarithms:  $log a - log b = log a/b$To multiply or divide, add or subtract logs: $log a + log b - log c =log ab/c$You can push a factor inside the log using:  $n log a = log a^n (n real)$logs of square roots simplify according to: $log \sqrt a = \onehalf  log a$logs of roots simplify according to: $log ^n\sqrt a = (1/n) log a$The log of 1 is 0.Factor a number completely to help simplify its logarithm.Factor out powers of 10 to help simplify the logarithm.Try writing $log(u)$ as $1/a log u^a$You could evaluate logs numerically.a logarithm in the exponent can be simplified by the law: $e^ln a = a$ln e = 1ln 1 = 0$ln e^n = n$ ($n$ real)You can write any power $u^v$  in the form  $e^(v ln u)$.a logarithm in the exponent can be simplified by the law: $e^((ln c) a) = c^a$$ln a^n = n ln a$.To multiply, add logarithms:  $ln ab = ln a + ln b$.The ln of a reciprical is the negative ln:  $ln 1/a = -ln a$.To divide, subtract lns:   $ln a/b = ln a - ln b$.Factor a number completely.Sums of natural logarithms combine according to: $ln a + ln b = ln ab$.Differences of natural logarithms combine according to: $ln a - ln b = ln a/b$.To multiply or divide, add or subtract natural logarithms: $ln a + ln b - ln c = ln (ab/c)$.$n ln a = ln a^n$  ($n$ real)natural logarithms of square roots simplify by:  $ln \sqrt a = \onehalf  ln a$.natural logarithms of roots simplify by: $ln ^n\sqrt a = (1/n) ln a$.Try writing $ln(1+v)$ as $v ln((1+v)^(1/v))$, and then use the limit definition of $e$Evaluate numerically.ln(a/b) = -ln(b/a)Use the formula for the sine of a sum in reverse.Use the formula for the sine of a difference in reverse.Use the formula for the cosine of a sum in reverse.Use the formula for the cosine of a difference in reverse.Use one of the formulas for the tangent of a half angle in reverse.Use one of the formulas for the cotangent of a half angle in reverse.Use the formula for the tangent of a sum in reverse.Use the formula for the tangent of a difference in reverse.Use the formula for the cotangent of a sum in reverse.Use the formula for the cotangent of a difference in reverse.Express $1 - cos \theta $ as $2 sin^2(\theta /2)$Express the complex number in polar formExpress the complex exponential using $sin$ and $cos$The complex exponential represents a point on the unit circle, which therefore has absolute value 1.The minus sign must be eliminated using $-a = ae^(i\pi )$.$^n\sqrt (-a)$ does not equal $-^n\sqrt a$ when complex numbers are in use.  Instead, a complex factor appears: $\sqrt (-a) = e^(\pi  i/n) ^n\sqrt a$.Complex exponents should be brought to the numerator.Use de Moivre's theorem, which gives a formula for the $n$ complex $n$-th roots of a number.Substitute specific integers for the integer parameter to obtain a complete list of specific solutions.Use the definition of logarithms: $b^(log(b,a)) = a$A logarithm in the exponent can be simplified by the law: $b^(n log(b,a)) = a^n$$log(b,b) = 1$$log(b,b^n) = n$A log of a product can be simplified using the law: $log xy = log x + log y$The log of a reciprocal can be simplified using the law: $log (1/x) = -log x$To divide, subtract logarithms:  $log x/y = log x-log y$$log(b,1) = 0$Factor the base of logarithms, for example, $log(4,x)=log(2^2,x)$$log(b^n,x) = (1/n) log (b,x)$$log x^n = n log x$Factor out powers of the base of logarithms.$log x + log y = log xy$$log x - log y = log x/y$$log x + log y - log z =log xy/z$$n log x = log x^n$ ($n$ real)Change the logarithms to natural logarithms.Change the logarithms to base 10.Change the base of the logarithms.Change the logs to a common base, using the law $log(b^n,x) = (1/n) log (b,x)$Base 10 logarithms can be written as logBase $e$ logarithms are written as lnChange log to lnChange ln to logExpress the power with the variable in the exponent, using $u^v = b^(v log(b,u))$.The sine function is zero at zero.The cosine function is one at zero.The tangent function is zero at zero.The zeroes of the sin function are at multiples of $\pi $cos takes the value 1 at even multiples of $\pi $The zeros of the tangent function are at multiples of $\pi $Since the trig functions are periodic, you should find a coterminal angle less than $360\deg $.  Select a trig function with an argument in the wrong range.Since the trig functions are periodic, you should find a coterminal angle less than $2\pi $.  Select a trig function with an argument in the wrong range.The values of the trig functions when the angle is a multiple of $90\deg $ are known.Use the relationships in a $1-2-\sqrt 3$ triangle.Use the relationships in a $1-1-\sqrt 2$ triangle.Change radians to degrees.Change degrees to radians.Express the angle in the form $a 30\deg  + b 45\deg $; then you can use sum formulas to break it up. Evaluate numericallyExpress tan in terms of sin and cosExpress cot in terms of tanExpress cot in terms of cos and sinExpress sec in terms of cosExpress csc in terms of sinCombine sin and cos into tanCombine cos and sin int cot%�|�4I:;I!I7I$>$>.@:;'I?	:;I
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Sindbad File Manager Version 1.0, Coded By Sindbad EG ~ The Terrorists