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The latest questions and answers, from members following Math at eNotes.Sat, 30 Aug 2014 03:14:47 PSTen-us`(x+2y+8z)-(6z-2y-x)` To subtract the expression inside the second...
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`(x+2y+8z)-(6z-2y-x)` To subtract the expression inside the second parenthesis from the expression inside the first parenthesis, consider the like terms present. These are x & -x, 2y & -2y, and 8z & 6z. To subtract like terms, subtract the numbers and copy the variable. `x-(-x)= x+x=(1+1)x=2x` `2y-(-2y)=2y+2y=(2+2)y=4y` `8z-6z=(8-6)z=2z` Then, write them together. Since all of them are positive, the operation between them would...http://www.enotes.com/homework-help/x-2y-8z-6z-2y-x-can-someone-explain-470693Sat, 30 Aug 2014 03:14:47 PST(x+2y+8z) - (6z-2y-x) Can someone explain?
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(x+2y+8z) - (6z-2y-x) Can someone explain?http://www.enotes.com/homework-help/x-2y-8z-6z-2y-x-can-someone-explain-470693#1August 30, 2014, 2:26 am PSTThe equation x^3+3*x^2+2*x = 0 has to be solved. x^3+3*x^2+2*x = 0 =>...
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The equation x^3+3*x^2+2*x = 0 has to be solved. x^3+3*x^2+2*x = 0 => x*(x^2+3*x+2) = 0 => x*(x^2 + 2x + x + 2) = 0 => x*(x(x + 2) + 1(x + 2)) = 0 => x*(x + 1)(x + 2) = 0 => x = 0, x = -1 and x = -2 The solution of the equation x^3+3*x^2+2*x = 0 is {-2, -1, 0}http://www.enotes.com/homework-help/solve-x-3-3-x-2-2-x-0-470687Fri, 29 Aug 2014 18:35:12 PSTSolve x^3+3*x^2+2*x = 0
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Solve x^3+3*x^2+2*x = 0http://www.enotes.com/homework-help/solve-x-3-3-x-2-2-x-0-470687#2August 29, 2014, 6:32 pm PSTAt the point of intersection of the curves defined by x^2 + 4y^2 = 16...
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At the point of intersection of the curves defined by x^2 + 4y^2 = 16 and y^2 - x^2 = 0 the x and y coordinates satisfy both the equations. y^2 - x^2 = 0 => x^2 = y^2 Substitute in x^2 + 4y^2 = 16 => 5x^2 = 16 => `x = +- 4/sqrt 5` The points of intersection of the given curves are `(4/sqrt 5, 4/sqrt 5)` ,`(4/sqrt 5,-4/sqrt 5)` ,`(-4/sqrt 5,4/sqrt 5)` and `(-4/sqrt 5,-4/sqrt 5)`http://www.enotes.com/homework-help/what-point-intersection-x-2-4y-2-16-y-2-x-2-0-470686Fri, 29 Aug 2014 18:31:52 PSTAlternatively, one can find the derivative of tan(x) by using the...
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Alternatively, one can find the derivative of tan(x) by using the identity `tan(x) = sin(x)/cos(x)` To find the derivative of a fraction, apply the quotient rule: `(f/g)' = (f'g - fg')/g^2` Here, f would be sin(x) and g would be cos(x). The derivatives of sine and cosine are (sin(x))' = cos(x) (cos(x))' = -sin(x) Substituting these functions into the quotient rule, we get `tan(x)' = (sin(x)/cos(x))' = (cos(x)cos(x) -...http://www.enotes.com/homework-help/prove-that-derivative-tan-x-sec-2-x-470682Fri, 29 Aug 2014 18:29:09 PSTWhat is the point of intersection of x^2 + 4y^2 = 16 and y^2 - x^2 = 0
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What is the point of intersection of x^2 + 4y^2 = 16 and y^2 - x^2 = 0http://www.enotes.com/homework-help/what-point-intersection-x-2-4y-2-16-y-2-x-2-0-470686#3August 29, 2014, 6:27 pm PSTThe curve y = 6x^2 + bx - 4 touches the x-axis if the equation 6x^2 + bx...
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The curve y = 6x^2 + bx - 4 touches the x-axis if the equation 6x^2 + bx - 4 = 0 has two equal roots. A quadratic equation ax^2 + bx + c has equal roots if b^2 = 4*a*c In the given problem, if the curve touches the x-axis the relation b^2 = -4*6*4 = -96 should hold. But this gives an imaginary value for b. The curve y = 6x^2 + bx - 4 cannot touch the x-axis for any value of bhttp://www.enotes.com/homework-help/when-does-curve-y-6x-2-bx-4-touch-x-axis-470685Fri, 29 Aug 2014 18:25:34 PSTWhen does the curve y = 6x^2 + bx - 4 touch the x-axis
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When does the curve y = 6x^2 + bx - 4 touch the x-axishttp://www.enotes.com/homework-help/when-does-curve-y-6x-2-bx-4-touch-x-axis-470685#4August 29, 2014, 6:21 pm PSTAt the points where a curve y = f(x) intersects the x-axis, the...
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At the points where a curve y = f(x) intersects the x-axis, the y-coordinate is equal to 0. To determine the points where y = 6x^2 + 7x - 4 intersects the x-axis solve y = 6x^2 + 7x - 4 = 0 => x = `(-7 +- sqrt(49 + 96))/12` => x = `(-7 +- sqrt 145)/12` The curve y = 6x^2 + 7x - 4 = 0 intersects the x-axis at points where the x-coordinate is `(-7 +- sqrt 145)/12`http://www.enotes.com/homework-help/what-values-x-does-y-6x-2-7x-4-touch-x-axis-470684Fri, 29 Aug 2014 18:19:43 PSTFor what values of x does y = 6x^2 + 7x - 4 intersect the x-axis.
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For what values of x does y = 6x^2 + 7x - 4 intersect the x-axis.http://www.enotes.com/homework-help/what-values-x-does-y-6x-2-7x-4-touch-x-axis-470684#5August 29, 2014, 6:14 pm PSTThe area of a regular pentagon with sides s is given by the formula `A =...
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The area of a regular pentagon with sides s is given by the formula `A = (1/4)*sqrt(5*(5+2*sqrt 5))*a^2` . For a regular pentagon with sides 4 cm the area is: `(1/4)*sqrt(5*(5+2*sqrt 5))*4^2` = `sqrt(5*(5+2*sqrt 5))*4` = `4*sqrt(5*(5+2*sqrt 5))` `~~ 27.527` cm^2 The area of a regular pentagon with sides 4 cm is approximately 27.527 cm^2http://www.enotes.com/homework-help/what-area-regular-pentagon-with-sides-4-cm-470683Fri, 29 Aug 2014 18:08:42 PSTWhat is the area of a regular pentagon with sides 4 cm.
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What is the area of a regular pentagon with sides 4 cm.http://www.enotes.com/homework-help/what-area-regular-pentagon-with-sides-4-cm-470683#6August 29, 2014, 6:04 pm PSTThe derivative of a function f(x) can be derived from first principles...
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The derivative of a function f(x) can be derived from first principles as the limit `lim_(h->0) (f(x+h) - f(x))/h` . For `f(x) = tan x = sin x/cos x` `f'(x) = lim_(h->0) (tan(x+h) - tan x)/h` = `lim_(h->0) ((sin(x+h))/(cos(x+h)) - sin x/cos x)/h` = `lim_(h->0) (sin(x+h)*cos x - cos(x+h)*sin x)/(cos(x+h)*cos x*h)` = `lim_(h->0) ((1/2)*sin(2x + h) + (1/2)*sin h - (1/2)*sin(2x + h) + (1/2)*sin h)/(cos(x+h)*cos x*h)` =...http://www.enotes.com/homework-help/prove-that-derivative-tan-x-sec-2-x-470682#7Fri, 29 Aug 2014 18:02:37 PSTProve that the derivative of tan x is sec^2 x.
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Prove that the derivative of tan x is sec^2 x.http://www.enotes.com/homework-help/prove-that-derivative-tan-x-sec-2-x-470682#8August 29, 2014, 5:46 pm PST