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The latest questions and answers, from members following Math at eNotes.Sun, 1 Mar 2015 12:51:33 PSTen-usThe number 1.0886621 has to be expressed in the form `(16*sqrt x)/9` ,...
http://www.enotes.com/homework-help/1-what-1-0886621-fraction-form-answer-wanted-form-475011
The number 1.0886621 has to be expressed in the form `(16*sqrt x)/9` , where x is a positive number. To determine x solve the equation `(16*sqrt x)/9 = 1.0886621` => `(256/81)*x = (1.0886621)^2` => `x = (1.0886621)^2*(81/256)` => `x ~~ .375 = 3/8` If x = 3/8, `(16*sqrt(3/8))/9 = 1.0886621` The point of inflection for a curve f(x) is the point where x = a and f''(a) = 0 Given `f(x) = (3x)/(x^2 - 9)` `f'(x) = 3/(x^2 - 9) -...http://www.enotes.com/homework-help/1-what-1-0886621-fraction-form-answer-wanted-form-475011Sun, 1 Mar 2015 12:51:33 PST1. What is 1.0886621 In fraction form. The answer is wanted in a form...
http://www.enotes.com/homework-help/1-what-1-0886621-fraction-form-answer-wanted-form-475011
1. What is 1.0886621 In fraction form. The answer is wanted in a form such as 16`sqrt(x)` /9 and I don't know how to find it quickly or find it at all. 2. Is the Point of Inflection for 3x/(x^2-9) coordinates (0,0)?http://www.enotes.com/homework-help/1-what-1-0886621-fraction-form-answer-wanted-form-475011#1March 1, 2015, 7:33 am PST`(x-4+i)(x-4-i)(x-3)` Tips to remember would be i x i = -1 The first...
http://www.enotes.com/homework-help/how-would-solve-this-x-4-x-4-x-3-128101
`(x-4+i)(x-4-i)(x-3)` Tips to remember would be i x i = -1 The first thing i would do are the first parentheses: `(x-4+i)(x-4-i)` and then foil: `x^2 - 4x - x i - 4x + 16 + 4i + x i - 4i -(-1)` move like terms next to each other to make it easier to combine: `x^2 - 4x - 4x - x i + x i + 4i - 4i + 16 + 1` simplify: `x^2 - 8x + 17` now multiply this by x - 3 `(x^2 - 8x + 17) (x - 3)` Foil: `x^3 - 3x^2 - 8x^2 + 24x + 17x - 51` combine like...http://www.enotes.com/homework-help/how-would-solve-this-x-4-x-4-x-3-128101Sat, 28 Feb 2015 23:30:25 PSTHow do you solve `5x^2-13x+6` ( `ax^2+bx+c` ) The first step of solving...
http://www.enotes.com/homework-help/how-do-you-solve-5x2-13x-6-also-how-do-you-solve-474109
How do you solve `5x^2-13x+6` ( `ax^2+bx+c` ) The first step of solving this is by multiplying a by c `5 xx 6 = 30` Then you find factors of 30 that add up to -13 (3 and 10) put those as b `5x^2 - 10x - 3x + 6` group `(5x^2 - 10x) (- 3x + 6)` Factor: `5x (x - 2) -3 (x - 2 )` `(5x - 3) (x - 2)` the solutions are x = 3/5 and x = 2 Also how do you solve x2-36 For this one we use the difference of 2 squares (a - b)(a + b) (x - 6) (x + 6)http://www.enotes.com/homework-help/how-do-you-solve-5x2-13x-6-also-how-do-you-solve-474109Sat, 28 Feb 2015 23:17:19 PSTy+6=2x 4x-10y=37 The first step is to solve one of the problems, y+6=2x...
http://www.enotes.com/homework-help/solve-each-system-using-substitution-y-6-2x-4x-10y-474195
y+6=2x 4x-10y=37 The first step is to solve one of the problems, y+6=2x seems easier to solve so i'll go with that first: y + 6 = 2x y = 2x - 6 Now plug in y as the y on the other problem: `4x - 10(2x-6) = 37` Now distribute the -10 `4x - 20x + 60 = 37` combine like terms `-16x = -23` divide by -16 x = 23/16 or 1.4375 ( the negative disappeared because both numbers are negative therefore cancelling it out) Now that we have x, plug it in as...http://www.enotes.com/homework-help/solve-each-system-using-substitution-y-6-2x-4x-10y-474195Sat, 28 Feb 2015 23:04:08 PSTSome iportant pointers I would give is that when something is the the...
http://www.enotes.com/homework-help/exponents-well-im-not-really-good-could-you-give-471751
Some iportant pointers I would give is that when something is the the 2nd power it is the same as multiplying the number by itself: `5^2 = 5 xx 5 = 25` A problem to the first power means that it is being multiplied by 5 `5^1 = 5 xx 1 = 5` To the 0 power is always 1, no matter what. A problem to the 0 power is 1 `5^0 = 1`http://www.enotes.com/homework-help/exponents-well-im-not-really-good-could-you-give-471751Sat, 28 Feb 2015 22:50:12 PSTThe limit `lim_(x-> oo) (x+7)/(3x+5)` has to be determined. If we...
http://www.enotes.com/homework-help/what-limit-x-7-3x-5-x-approaches-infinity-255867
The limit `lim_(x-> oo) (x+7)/(3x+5)` has to be determined. If we substitute `x = oo` in `(x+7)/(3x+5)` we get the indeterminate form `oo/oo` , To determine the limit in this case we can use l'Hospital's rule and substitute the numerator and denominator by their derivatives. (x +7)' = 1 (3x + 5)' = 3 This gives the limit `lim_(x->oo)1/3` As the variable x does not figure in `1/3` , this is the required limit. The limit `lim_(x-> oo)...http://www.enotes.com/homework-help/what-limit-x-7-3x-5-x-approaches-infinity-255867Sat, 28 Feb 2015 20:06:33 PSTThe limit `lim_(x->-2)(x^5+32)/(x+2)` has to be determined. If we...
http://www.enotes.com/homework-help/limit-x-5-32-x-2-x-gt-2-298112
The limit `lim_(x->-2)(x^5+32)/(x+2)` has to be determined. If we substitute x = -2 in `(x^5+32)/(x+2)` the result is `(-2^5 + 32)/(-2+2) = 0/0 ` which is indeterminate. This is also i a form that allows us to use l'Hospital's rule and replace the numerator and denominator by their derivatives. `(x^5 + 32)' = 5x^4` `(x + 2)' = 1` The limit can be written as: `lim_(x -> -2) (5x^4)/1` Substituting x = -2 gives 5*16 = 80 The required limit...http://www.enotes.com/homework-help/limit-x-5-32-x-2-x-gt-2-298112Sat, 28 Feb 2015 19:58:55 PSTThe value of the definite integral `int_0^5 2x^3 - 4x^2 + 1 dx` has to...
http://www.enotes.com/homework-help/evaluate-integral-upper-limit-5-lower-limit-0-2x-3-336611
The value of the definite integral `int_0^5 2x^3 - 4x^2 + 1 dx` has to be determined. Use the rule `int x^n dx = (x^(n+1))/(n+1)` `int_0^5 2x^3 - 4x^2 + 1 dx` = `int_0^5 2x^3 dx - int_0^5 4x^2 dx + int_0^5 1 dx` = `[(2*x^4)/4]_0^5 - [4*x^3/3]_0^5 + [x]_0^5` = `5^4/2 - (4/3)*5^3 + 5` = `625/2 - 500/3 + 5 ` = `905/6` The required value of `int_0^5 2x^3 - 4x^2 + 1 dx = 905/6`http://www.enotes.com/homework-help/evaluate-integral-upper-limit-5-lower-limit-0-2x-3-336611Sat, 28 Feb 2015 19:53:40 PSTThe value of the limit `lim_(h->0) (sin(pi/2+h) - sin (pi/2))/h` has...
http://www.enotes.com/homework-help/evaluate-without-using-calculator-limit-h-316152
The value of the limit `lim_(h->0) (sin(pi/2+h) - sin (pi/2))/h` has to be determined. The easiest way to determine the required value is by keeping in mind that the derivative of any function f(x) from first principles is the limit `lim_(h->0)(f(x+h) - f(x))/h` The limit `lim_(h->0) (sin(pi/2+h) - sin (pi/2))/h` is the derivative of f(x) = sin (x) for `x = pi/2` . The value of the limit is `(sin(pi/2)' = cos(pi/2) = 0` The required...http://www.enotes.com/homework-help/evaluate-without-using-calculator-limit-h-316152Sat, 28 Feb 2015 19:46:42 PSTThe limit `lim_(h->0)(sqrt(81+h) - 9)/h` is required. Substituting h...
http://www.enotes.com/homework-help/limit-lim-h-gt-0-sqrt-81-h-9-hrepresents-315862
The limit `lim_(h->0)(sqrt(81+h) - 9)/h` is required. Substituting h = 0 in the expression `(sqrt(81+h) - 9)/h` gives `(sqrt(81+0) - 9)/0 = 0/0` which is indeterminate. We have to change the form of the expression to determine the limit. `lim_(h->0)(sqrt(81+h) - 9)/h` = `lim_(h->0)(sqrt(81+h) - 9)/(81 + h - 81)` = `lim_(h->0)(sqrt(81+h) - 9)/((sqrt(81 + h)^2 - 9^2)` Use the relation x^2 - y^2 = (x + y)(x - y) =...http://www.enotes.com/homework-help/limit-lim-h-gt-0-sqrt-81-h-9-hrepresents-315862Sat, 28 Feb 2015 19:40:40 PSTThe definite integral `int_0^1 x^2+x dx` has to be determined. The...
http://www.enotes.com/homework-help/calculate-definite-integral-y-x-2-x-x-0-x-1-267966
The definite integral `int_0^1 x^2+x dx` has to be determined. The derivative of x^n is `(x^(n+1))/(n+1)` . Using this rule, the given integral can be written as: `int_0^1 x^2+x dx` = `[x^3/3 + x^2/2]_0^1` = `1/3 - 0 + 1/2 - 0` = `5/6` The required definite integral `int_0^1 x^2+x dx = 5/6`http://www.enotes.com/homework-help/calculate-definite-integral-y-x-2-x-x-0-x-1-267966Sat, 28 Feb 2015 19:34:20 PSTThe limit `lim_(x->6)(x^3 - 216)/(x - 6)` has to be determined....
http://www.enotes.com/homework-help/what-limit-x-6-x-3-216-x-6-336566
The limit `lim_(x->6)(x^3 - 216)/(x - 6)` has to be determined. Substituting x = 6 in the given expression gives the result `(6^3 - 216)/(6-6) = 0/0` which is indeterminate. The factorized form of `x^3 - y^3 = (x - y)*(y^2+x*y+x^2)` . We can rewrite the limit as: `lim_(x->6) ((x - 6)(x^2 + 6x + 6^2)/(x - 6)` = `lim_(x->6) (x^2 + 6x + 6^2)` Substituting x = 6 gives 36 + 36 + 36 = 108 The required limit `lim_(x->6)(x^3 - 216)/(x - 6)...http://www.enotes.com/homework-help/what-limit-x-6-x-3-216-x-6-336566Sat, 28 Feb 2015 19:26:34 PSTFor a function f(x), the limit `lim_(h->0)(f(x+h) - f(x))/h` is the...
http://www.enotes.com/homework-help/limits-question-330924
For a function f(x), the limit `lim_(h->0)(f(x+h) - f(x))/h` is the derivative f'(x) of the function. Here, we are supposed to determine `lim_(h->0) (sin(pi+h) - sin pi)/h` `lim_(h->0) (sin(pi+h) - sin pi)/h` is the derivative of sin x for `x = pi` . There is no need to actually calculate the value of the limit, use the derivative of the function sin x which is cos x. This gives `lim_(h->0) (sin(pi+h) - sin pi)/h = cos pi = -1` The...http://www.enotes.com/homework-help/limits-question-330924Sat, 28 Feb 2015 19:20:06 PSTFor a function f(x), the limit `lim_(h->0) (f(x+h) - f(x))/h` is the...
http://www.enotes.com/homework-help/given-f-x-2-3x-4-find-lim-h-gt-0-f-2-h-f-2-h-343731
For a function f(x), the limit `lim_(h->0) (f(x+h) - f(x))/h` is the derivative f'(x) of the function. Here, we have `f(x) = 2/(3x-4)` and we are required to find the limit `lim_(h->0)(f(2+h)-f(2))/h` . `lim_(h->0)(f(2+h)-f(2))/h` is equivalent to `f'(x)` for x = 2. As `f(x) = 2/(3x - 4)` `f'(x) = 2*-1*3/(3x - 4)^2` `f'(2) = -6/(6 - 4)^2` `= -6/4` `= -3/2` The required limit `lim_(h->0)(f(2+h)-f(2))/h = -3/2`http://www.enotes.com/homework-help/given-f-x-2-3x-4-find-lim-h-gt-0-f-2-h-f-2-h-343731Sat, 28 Feb 2015 19:14:15 PST