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FACTORING POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
MAKE PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Dividing x^2 by x2 + x - 6, we obtain. For the fraction (x - 6)/ (x^2 + x - 6), we now find a partial fraction decomposition as follows. x - 6 = A (x - 2) + B (x + 3). To find A and B, we choose x = 2 and x = - 3 and obtain the following results. To check this, simply add the fractions on the right. The result should be the fraction on the left THE OUTCOME OF EXPANDING (Q-8)*(Q+3) STEP-BY-STEP MATH Result. A sum containing 3 terms. The 1st term of the sum is equal to a power. The base is q. The exponent is two. The 2nd term of the sum is equal to a negative product containing 2 factors. The 1st factor of the product is equal to five. The 2nd factor of the product is q. The 3rd term of the sum is equal to negative twenty four. STEP-BY-STEP MATH PROBLEM SOLVERALGEBRAEQUATIONSINEQUALITIESGRAPHSNUMBERSCALCULUS QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and SIMPLIFYING RADICALS STEP-BY-STEP MATH PROBLEM SOLVER Examples. Simplify the following radicals. 1. √24 Factor 24 so that one factor is a square number. √24=√4·6=√4·√6=2 √6. 2. √72 Find the largest square factor you can before simplifying. √72=√36·2=√36·√2=6 √2. Or, if you did not notice 36 as a factor, you could write. √72=√9·8=√9·√8=3 √4·2=3·√4·√2=3 FACTORING STEP-BY-STEP MATH PROBLEM SOLVER Example 6. Factor a^4-5a^2+4. Although this is not a quadratic in a, it is a quadratic in a^2, that is, a^4-5a^2+4=(a^2)^2-5(a^2)+4 Since the coefficient of a^2 is negative, the two possible factorization of 4 are (-4) (-1) and (-2) (-2).Since (-4) + (-1) = -5 5, we have a^4-5a^2+4=(a^2+4)(a^2-1) Since each of the factors obtained is the difference of two squares we factor again obtaining RATIONAL EXPRESSIONS STEP-BY-STEP MATH PROBLEM SOLVER An expression that is the quotient of two algebraic expressions (with denominator not 0) is called a fractional expression. The most common fractional expressions are those that are the quotients of two polynomials; these are called rational expressions. GRAPH EQUATIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER The point-slope form of a line with slope m and passing through the point (x 1, y 1 ) is. y - y 1 - m (x - x 1) The slope-intercept form of a line with slope m and y-intercept b is. y = mx + b. A relationship determined by an equation of the form. y = kx (k a constant) is called a direct variation. COORDINATES IN THE PLANE & GRAPHING... STEP-BY-STEP MATH 7.1 Coordinates in the Plane. Let PI be a plane and let X and Y be mutually perpendicular lines in PI intersecting at the point O.Using the lines X and Y we will associate a number pair with each point in the plane. If P is a point and (a,b) is the pair associated with P, then a and b are the coordinates of P.The number a is the abcissa or first coordinate of P. while b is the ordinate or SYSTEM OF EQUATIONS STEP-BY-STEP MATH PROBLEM SOLVER The numbers a1,b1,a2 and b2 are called the elements of the determinant. For example, We will now show how determinants can be used to solve the system (1) above. a1 x+b1 y=c1. a2 x+b2 y=c2. Multiply the first equation by , the second equation by − b1, andadd to
FACTORING POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
MAKE PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Dividing x^2 by x2 + x - 6, we obtain. For the fraction (x - 6)/ (x^2 + x - 6), we now find a partial fraction decomposition as follows. x - 6 = A (x - 2) + B (x + 3). To find A and B, we choose x = 2 and x = - 3 and obtain the following results. To check this, simply add the fractions on the right. The result should be the fraction on the left THE OUTCOME OF EXPANDING (Q-8)*(Q+3) STEP-BY-STEP MATH Result. A sum containing 3 terms. The 1st term of the sum is equal to a power. The base is q. The exponent is two. The 2nd term of the sum is equal to a negative product containing 2 factors. The 1st factor of the product is equal to five. The 2nd factor of the product is q. The 3rd term of the sum is equal to negative twenty four. GRAPH EQUATIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER The point-slope form of a line with slope m and passing through the point (x 1, y 1 ) is. y - y 1 - m (x - x 1) The slope-intercept form of a line with slope m and y-intercept b is. y = mx + b. A relationship determined by an equation of the form. y = kx (k a constant) is called a direct variation. POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVER Arrange both polynomials in descending powers of x, and write as follows. Step 2. Divide x into 2 x2. Step 3. Multiply x-1 by 2 x and subtract from 2 x2-3 x+5. Since the degree of the remainder, − x+5, is not less than the degree of the divisor, x-1, we repeat theprocess. Step 4.
MAKE PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Dividing x^2 by x2 + x - 6, we obtain. For the fraction (x - 6)/ (x^2 + x - 6), we now find a partial fraction decomposition as follows. x - 6 = A (x - 2) + B (x + 3). To find A and B, we choose x = 2 and x = - 3 and obtain the following results. To check this, simply add the fractions on the right. The result should be the fraction on the left SOLVE INEQUALITIES WITH STEP-BY-STEP MATH PROBLEM SOLVER Example 1 In the formula d = rt, find t if d = 24 and r = 3. Solution We can solve for t by substituting 24 for d and 3 for r. That is, d = rt. (24) = (3)t. 8 = t. It is often necessary to solve formulas or equations in which there is more than one variable for one of the variables in terms of the others. DIFFERENTIATE A FUNCTION WITH STEP-BY-STEP MATH PROBLEM SOLVER Differentiate each of the following functions: (a) Since f (x) = 5, f is a constant function; hence f ' (x) = 0. (b) With n = 15 in the power rule, f ' (x) = 15x 14. (c) Note that f (x) = x 1/2 . Hence, with n = 1/2 in the power rule, (d) Since f (x) = x -1, it follows from the power rule that f ' (x) = -x -2 = -1/x 2. EQUATION OF CIRCLE AND PARABOLAS STEP-BY-STEP MATH PROBLEM 7.5 The Equation of a Circle A circle C in the XY plane, with center at the point (h, k) and radius r, is the set of all points at distance r from the point (h, k).Let P: (x,y) be any point on C.Then by the distance formula from Section 7.1 we have root((x-h)^2+(y-k)^2)=r An equivalent equation is DIVISION OF POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVER Solve an equation, inequality or a system. Example: 2x-1=y,2y+3=x To see your tutorial, please scroll down RATIONAL FORMS STEP-BY-STEP MATH PROBLEM SOLVER CHAPTER 4 Rational Forms. 4.1 Introduction By a rational form we will mean an algebraic expression of the form P/Q, where P and Q are polynomials in one or more variables, and Q is nonzero. The polynomial P will be called the numerator of the rational form and Q will be called the denominator. In Chapter 1, we pointed out that if a,b, and c are any three real numbers with b!=0, and c!=0, then FIND MATRIX INVERSE WITH STEP-BY-STEP MATH PROBLEM SOLVER Finally, The process for finding the multiplicative inverse A^ (-1) n x n matrix A that has an inverse is summarized below. FINDING AN INVERSE MATRIX. To obtain A^ (-1) n x n matrix A for which A^ (-1) exists, follow these steps. 1. Form the augmented matrix , where I is the n x n identity matrix. X*(X-5) = 0 EQUATION SOLVED STEP-BY-STEP MATH PROBLEM SOLVER The first factor of the product is equal to x. The second factor of the product is equal to a sum of 2 terms. The first term of the sum is equal to x. The second term of the sum is equal to negative five Right side of the equation is equal to zero. x times left brace x plus negative five right brace is equal to zero; Variable. STEP-BY-STEP MATH PROBLEM SOLVERALGEBRAEQUATIONSINEQUALITIESGRAPHSNUMBERSCALCULUS QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and SIMPLIFYING RADICALS STEP-BY-STEP MATH PROBLEM SOLVER Examples. Simplify the following radicals. 1. √24 Factor 24 so that one factor is a square number. √24=√4·6=√4·√6=2 √6. 2. √72 Find the largest square factor you can before simplifying. √72=√36·2=√36·√2=6 √2. Or, if you did not notice 36 as a factor, you could write. √72=√9·8=√9·√8=3 √4·2=3·√4·√2=3 FACTORING STEP-BY-STEP MATH PROBLEM SOLVER Example 6. Factor a^4-5a^2+4. Although this is not a quadratic in a, it is a quadratic in a^2, that is, a^4-5a^2+4=(a^2)^2-5(a^2)+4 Since the coefficient of a^2 is negative, the two possible factorization of 4 are (-4) (-1) and (-2) (-2).Since (-4) + (-1) = -5 5, we have a^4-5a^2+4=(a^2+4)(a^2-1) Since each of the factors obtained is the difference of two squares we factor again obtaining RATIONAL EXPRESSIONS STEP-BY-STEP MATH PROBLEM SOLVER An expression that is the quotient of two algebraic expressions (with denominator not 0) is called a fractional expression. The most common fractional expressions are those that are the quotients of two polynomials; these are called rational expressions. GRAPH EQUATIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER The point-slope form of a line with slope m and passing through the point (x 1, y 1 ) is. y - y 1 - m (x - x 1) The slope-intercept form of a line with slope m and y-intercept b is. y = mx + b. A relationship determined by an equation of the form. y = kx (k a constant) is called a direct variation. COORDINATES IN THE PLANE & GRAPHING... STEP-BY-STEP MATH 7.1 Coordinates in the Plane. Let PI be a plane and let X and Y be mutually perpendicular lines in PI intersecting at the point O.Using the lines X and Y we will associate a number pair with each point in the plane. If P is a point and (a,b) is the pair associated with P, then a and b are the coordinates of P.The number a is the abcissa or first coordinate of P. while b is the ordinate or SYSTEM OF EQUATIONS STEP-BY-STEP MATH PROBLEM SOLVER The numbers a1,b1,a2 and b2 are called the elements of the determinant. For example, We will now show how determinants can be used to solve the system (1) above. a1 x+b1 y=c1. a2 x+b2 y=c2. Multiply the first equation by , the second equation by − b1, andadd to
FACTORING POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
MAKE PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Dividing x^2 by x2 + x - 6, we obtain. For the fraction (x - 6)/ (x^2 + x - 6), we now find a partial fraction decomposition as follows. x - 6 = A (x - 2) + B (x + 3). To find A and B, we choose x = 2 and x = - 3 and obtain the following results. To check this, simply add the fractions on the right. The result should be the fraction on the left THE OUTCOME OF EXPANDING (Q-8)*(Q+3) STEP-BY-STEP MATH Result. A sum containing 3 terms. The 1st term of the sum is equal to a power. The base is q. The exponent is two. The 2nd term of the sum is equal to a negative product containing 2 factors. The 1st factor of the product is equal to five. The 2nd factor of the product is q. The 3rd term of the sum is equal to negative twenty four. STEP-BY-STEP MATH PROBLEM SOLVERALGEBRAEQUATIONSINEQUALITIESGRAPHSNUMBERSCALCULUS QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and SIMPLIFYING RADICALS STEP-BY-STEP MATH PROBLEM SOLVER Examples. Simplify the following radicals. 1. √24 Factor 24 so that one factor is a square number. √24=√4·6=√4·√6=2 √6. 2. √72 Find the largest square factor you can before simplifying. √72=√36·2=√36·√2=6 √2. Or, if you did not notice 36 as a factor, you could write. √72=√9·8=√9·√8=3 √4·2=3·√4·√2=3 FACTORING STEP-BY-STEP MATH PROBLEM SOLVER Example 6. Factor a^4-5a^2+4. Although this is not a quadratic in a, it is a quadratic in a^2, that is, a^4-5a^2+4=(a^2)^2-5(a^2)+4 Since the coefficient of a^2 is negative, the two possible factorization of 4 are (-4) (-1) and (-2) (-2).Since (-4) + (-1) = -5 5, we have a^4-5a^2+4=(a^2+4)(a^2-1) Since each of the factors obtained is the difference of two squares we factor again obtaining RATIONAL EXPRESSIONS STEP-BY-STEP MATH PROBLEM SOLVER An expression that is the quotient of two algebraic expressions (with denominator not 0) is called a fractional expression. The most common fractional expressions are those that are the quotients of two polynomials; these are called rational expressions. GRAPH EQUATIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER The point-slope form of a line with slope m and passing through the point (x 1, y 1 ) is. y - y 1 - m (x - x 1) The slope-intercept form of a line with slope m and y-intercept b is. y = mx + b. A relationship determined by an equation of the form. y = kx (k a constant) is called a direct variation. COORDINATES IN THE PLANE & GRAPHING... STEP-BY-STEP MATH 7.1 Coordinates in the Plane. Let PI be a plane and let X and Y be mutually perpendicular lines in PI intersecting at the point O.Using the lines X and Y we will associate a number pair with each point in the plane. If P is a point and (a,b) is the pair associated with P, then a and b are the coordinates of P.The number a is the abcissa or first coordinate of P. while b is the ordinate or SYSTEM OF EQUATIONS STEP-BY-STEP MATH PROBLEM SOLVER The numbers a1,b1,a2 and b2 are called the elements of the determinant. For example, We will now show how determinants can be used to solve the system (1) above. a1 x+b1 y=c1. a2 x+b2 y=c2. Multiply the first equation by , the second equation by − b1, andadd to
FACTORING POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
MAKE PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Dividing x^2 by x2 + x - 6, we obtain. For the fraction (x - 6)/ (x^2 + x - 6), we now find a partial fraction decomposition as follows. x - 6 = A (x - 2) + B (x + 3). To find A and B, we choose x = 2 and x = - 3 and obtain the following results. To check this, simply add the fractions on the right. The result should be the fraction on the left THE OUTCOME OF EXPANDING (Q-8)*(Q+3) STEP-BY-STEP MATH Result. A sum containing 3 terms. The 1st term of the sum is equal to a power. The base is q. The exponent is two. The 2nd term of the sum is equal to a negative product containing 2 factors. The 1st factor of the product is equal to five. The 2nd factor of the product is q. The 3rd term of the sum is equal to negative twenty four. GRAPH EQUATIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER The point-slope form of a line with slope m and passing through the point (x 1, y 1 ) is. y - y 1 - m (x - x 1) The slope-intercept form of a line with slope m and y-intercept b is. y = mx + b. A relationship determined by an equation of the form. y = kx (k a constant) is called a direct variation. POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVER Arrange both polynomials in descending powers of x, and write as follows. Step 2. Divide x into 2 x2. Step 3. Multiply x-1 by 2 x and subtract from 2 x2-3 x+5. Since the degree of the remainder, − x+5, is not less than the degree of the divisor, x-1, we repeat theprocess. Step 4.
MAKE PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Dividing x^2 by x2 + x - 6, we obtain. For the fraction (x - 6)/ (x^2 + x - 6), we now find a partial fraction decomposition as follows. x - 6 = A (x - 2) + B (x + 3). To find A and B, we choose x = 2 and x = - 3 and obtain the following results. To check this, simply add the fractions on the right. The result should be the fraction on the left SOLVE INEQUALITIES WITH STEP-BY-STEP MATH PROBLEM SOLVER Example 1 In the formula d = rt, find t if d = 24 and r = 3. Solution We can solve for t by substituting 24 for d and 3 for r. That is, d = rt. (24) = (3)t. 8 = t. It is often necessary to solve formulas or equations in which there is more than one variable for one of the variables in terms of the others. DIFFERENTIATE A FUNCTION WITH STEP-BY-STEP MATH PROBLEM SOLVER Differentiate each of the following functions: (a) Since f (x) = 5, f is a constant function; hence f ' (x) = 0. (b) With n = 15 in the power rule, f ' (x) = 15x 14. (c) Note that f (x) = x 1/2 . Hence, with n = 1/2 in the power rule, (d) Since f (x) = x -1, it follows from the power rule that f ' (x) = -x -2 = -1/x 2. EQUATION OF CIRCLE AND PARABOLAS STEP-BY-STEP MATH PROBLEM 7.5 The Equation of a Circle A circle C in the XY plane, with center at the point (h, k) and radius r, is the set of all points at distance r from the point (h, k).Let P: (x,y) be any point on C.Then by the distance formula from Section 7.1 we have root((x-h)^2+(y-k)^2)=r An equivalent equation is DIVISION OF POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVER Solve an equation, inequality or a system. Example: 2x-1=y,2y+3=x To see your tutorial, please scroll down RATIONAL FORMS STEP-BY-STEP MATH PROBLEM SOLVER CHAPTER 4 Rational Forms. 4.1 Introduction By a rational form we will mean an algebraic expression of the form P/Q, where P and Q are polynomials in one or more variables, and Q is nonzero. The polynomial P will be called the numerator of the rational form and Q will be called the denominator. In Chapter 1, we pointed out that if a,b, and c are any three real numbers with b!=0, and c!=0, then FIND MATRIX INVERSE WITH STEP-BY-STEP MATH PROBLEM SOLVER Finally, The process for finding the multiplicative inverse A^ (-1) n x n matrix A that has an inverse is summarized below. FINDING AN INVERSE MATRIX. To obtain A^ (-1) n x n matrix A for which A^ (-1) exists, follow these steps. 1. Form the augmented matrix , where I is the n x n identity matrix. X*(X-5) = 0 EQUATION SOLVED STEP-BY-STEP MATH PROBLEM SOLVER The first factor of the product is equal to x. The second factor of the product is equal to a sum of 2 terms. The first term of the sum is equal to x. The second term of the sum is equal to negative five Right side of the equation is equal to zero. x times left brace x plus negative five right brace is equal to zero; Variable. STEP-BY-STEP MATH PROBLEM SOLVERALGEBRAEQUATIONSINEQUALITIESGRAPHSNUMBERSCALCULUS QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and SOLVE LINEAR, HIGHER ORDER EQUATIONS WITH STEP-BY-STEPSEE MORE ONQUICKMATH.COM
RATIONAL EXPRESSIONS STEP-BY-STEP MATH PROBLEM SOLVER An expression that is the quotient of two algebraic expressions (with denominator not 0) is called a fractional expression. The most common fractional expressions are those that are the quotients of two polynomials; these are called rational expressions. EXPAND TERMS, MULTIPLY POLYNOMIALS WITH STEP-BY-STEP MATH OBJECTIVES. Find the product of two binomials. Use the distributive property to multiply any two polynomials. In the previous section you learned that the product A (2x + y) expands to A (2x) + A (y). Now consider the product (3x + z) (2x + y). Since (3x + z) is in parentheses, we can treat it as a single factor and expand (3x + z)(2x + y) in
INTEGRATE A FUNCTION WITH STEP-BY-STEP MATH PROBLEM SOLVER Find the corresponding cost function C (x). We have already seen that any cost function for this marginal cost must be of the form C (x) = x 2 + a for some constant a. Since. C (0) = 500 = 0 2 + a = a, we have a = 500. Thus, the cost function is given by C (x) = x 2 + 500. From this example, we see that the arbitrary constant c is the fixed DIFFERENTIATE A FUNCTION WITH STEP-BY-STEP MATH PROBLEM SOLVER Differentiate each of the following functions: (a) Since f (x) = 5, f is a constant function; hence f ' (x) = 0. (b) With n = 15 in the power rule, f ' (x) = 15x 14. (c) Note that f (x) = x 1/2 . Hence, with n = 1/2 in the power rule, (d) Since f (x) = x -1, it follows from the power rule that f ' (x) = -x -2 = -1/x 2. SOLVE INEQUALITIES WITH STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
FACTORING POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
FIND PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Example 1. Solution Note that the denominator of the integrand can be factored: The plan is to decompose this fraction into partial fractions by finding numbers A and B for which. holds for all x except x = 1 and x = - 2. If this is possible, then we can integrate 1/ (x^2+x-2) by finding : since these last two antiderivatives can beevaluated
FACTOR A POLYNOMIAL OR AN EXPRESSION WITH STEP-BY-STEPSEE MORE ONQUICKMATH.COM
STEP-BY-STEP MATH PROBLEM SOLVERALGEBRAEQUATIONSINEQUALITIESGRAPHSNUMBERSCALCULUS QuickMath will automatically answer the most common problems in algebra, equations and calculus faced by high-school and college students. The algebra section allows you to expand, factor or simplify virtually any expression you choose. It also has commands for splitting fractions into partial fractions, combining several fractions into one and SOLVE LINEAR, HIGHER ORDER EQUATIONS WITH STEP-BY-STEPSEE MORE ONQUICKMATH.COM
RATIONAL EXPRESSIONS STEP-BY-STEP MATH PROBLEM SOLVER An expression that is the quotient of two algebraic expressions (with denominator not 0) is called a fractional expression. The most common fractional expressions are those that are the quotients of two polynomials; these are called rational expressions. EXPAND TERMS, MULTIPLY POLYNOMIALS WITH STEP-BY-STEP MATH OBJECTIVES. Find the product of two binomials. Use the distributive property to multiply any two polynomials. In the previous section you learned that the product A (2x + y) expands to A (2x) + A (y). Now consider the product (3x + z) (2x + y). Since (3x + z) is in parentheses, we can treat it as a single factor and expand (3x + z)(2x + y) in
INTEGRATE A FUNCTION WITH STEP-BY-STEP MATH PROBLEM SOLVER Find the corresponding cost function C (x). We have already seen that any cost function for this marginal cost must be of the form C (x) = x 2 + a for some constant a. Since. C (0) = 500 = 0 2 + a = a, we have a = 500. Thus, the cost function is given by C (x) = x 2 + 500. From this example, we see that the arbitrary constant c is the fixed DIFFERENTIATE A FUNCTION WITH STEP-BY-STEP MATH PROBLEM SOLVER Differentiate each of the following functions: (a) Since f (x) = 5, f is a constant function; hence f ' (x) = 0. (b) With n = 15 in the power rule, f ' (x) = 15x 14. (c) Note that f (x) = x 1/2 . Hence, with n = 1/2 in the power rule, (d) Since f (x) = x -1, it follows from the power rule that f ' (x) = -x -2 = -1/x 2. SOLVE INEQUALITIES WITH STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
FACTORING POLYNOMIALS STEP-BY-STEP MATH PROBLEM SOLVERSEE MORE ONQUICKMATH.COM
FIND PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Example 1. Solution Note that the denominator of the integrand can be factored: The plan is to decompose this fraction into partial fractions by finding numbers A and B for which. holds for all x except x = 1 and x = - 2. If this is possible, then we can integrate 1/ (x^2+x-2) by finding : since these last two antiderivatives can beevaluated
FACTOR A POLYNOMIAL OR AN EXPRESSION WITH STEP-BY-STEPSEE MORE ONQUICKMATH.COM
MATH ARTICLES STEP-BY-STEP MATH PROBLEM SOLVER Quadratic equations. Quadratic equations are a little bit trickier than linear ones. If you are unsure about linear equations go over linear equations tutorials first. If you think you can handle them, use our quadratic equations tutorials and interactive solvers to learn about quadratic formula and other ways of solving second orderequations.
FACTORING STEP-BY-STEP MATH PROBLEM SOLVER Example 6. Factor a^4-5a^2+4. Although this is not a quadratic in a, it is a quadratic in a^2, that is, a^4-5a^2+4=(a^2)^2-5(a^2)+4 Since the coefficient of a^2 is negative, the two possible factorization of 4 are (-4) (-1) and (-2) (-2).Since (-4) + (-1) = -5 5, we have a^4-5a^2+4=(a^2+4)(a^2-1) Since each of the factors obtained is the difference of two squares we factor again obtaining GRAPH EQUATIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER The point-slope form of a line with slope m and passing through the point (x 1, y 1 ) is. y - y 1 - m (x - x 1) The slope-intercept form of a line with slope m and y-intercept b is. y = mx + b. A relationship determined by an equation of the form. y = kx (k a constant) is called a direct variation. INTEGRATE A FUNCTION WITH STEP-BY-STEP MATH PROBLEM SOLVER Find the corresponding cost function C (x). We have already seen that any cost function for this marginal cost must be of the form C (x) = x 2 + a for some constant a. Since. C (0) = 500 = 0 2 + a = a, we have a = 500. Thus, the cost function is given by C (x) = x 2 + 500. From this example, we see that the arbitrary constant c is the fixed SIMPLIFYING RADICALS STEP-BY-STEP MATH PROBLEM SOLVER Examples. Simplify the following radicals. 1. √24 Factor 24 so that one factor is a square number. √24=√4·6=√4·√6=2 √6. 2. √72 Find the largest square factor you can before simplifying. √72=√36·2=√36·√2=6 √2. Or, if you did not notice 36 as a factor, you could write. √72=√9·8=√9·√8=3 √4·2=3·√4·√2=3 FIND PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Example 1. Solution Note that the denominator of the integrand can be factored: The plan is to decompose this fraction into partial fractions by finding numbers A and B for which. holds for all x except x = 1 and x = - 2. If this is possible, then we can integrate 1/ (x^2+x-2) by finding : since these last two antiderivatives can beevaluated
SIMPLIFY RADICAL,RATIONAL EXPRESSION WITH STEP-BY-STEP That is the reason the x 3 term was missing or not written in the original expression. Solution. Step 1: Arrange both the divisor and dividend in descending powers of the variable (this means highest exponent first, next highest second, and so on) and supply a zero coefficient for any missing terms. SOLVE INEQUALITIES WITH STEP-BY-STEP MATH PROBLEM SOLVER Example 5 is a formula giving interest (I) earned for a period of D days when the principal (p) and the yearly rate (r) are known. Find the yearly rate when the amount of interest, the principal, and the number of days are all known. Solution. The problem requires solving for r.. Notice in this example that r was left on the right side and thus the computation was simpler. MAKE PARTIAL FRACTIONS WITH STEP-BY-STEP MATH PROBLEM SOLVER Dividing x^2 by x2 + x - 6, we obtain. For the fraction (x - 6)/ (x^2 + x - 6), we now find a partial fraction decomposition as follows. x - 6 = A (x - 2) + B (x + 3). To find A and B, we choose x = 2 and x = - 3 and obtain the following results. To check this, simply add the fractions on the right. The result should be the fraction on the left GRAPH INEQUALITIES WITH STEP-BY-STEP MATH PROBLEM SOLVER Solution Step 1: First graph 2x - y = 4. Since the line graph for 2x - y = 4 does not go through the origin (0,0), check that point in the linear inequality. Step 2: Step 3: Since the point (0,0) is not in the solution set, the half-plane containing (0,0) is not in the set. Hence, the solution is the other half-plane. HOME | ABOUT | CONTACT| DISCLAIMER
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