dyke 270 Posted March 7, 2014 Share Posted March 7, 2014 if u need help with ur hw, then feel free to comment on this thread ) so yea i'll be the first one i need help with this problem: log2(y+4)+ log2y=5 how is it 4 and can som1 tell me how to determine if its extraneous or we thank u i will donate 100k won if u helped me @!!!!!!!! (if i reallly understand it) Link to comment Share on other sites More sharing options...
Leo's Candy Jelly Ho 3,458 Posted March 7, 2014 Share Posted March 7, 2014 Use implicit differentiation to find dy/dx 3y^2= (4x-5)/(4x+5) dy/dx Link to comment Share on other sites More sharing options...
Ghettoe 1,955 Posted March 7, 2014 Share Posted March 7, 2014 I sucked at math, Use implicit differentiation to find dy/dx 3y^2= (4x-5)/(4x+5) dy/dx EDIT: I HAVE REMEMBERED. *face palm* 2(3y)^(2-1) = (4x-5)(4) - (4x+5)(4)/ (4x+5)^2 = (16x-20) - (16x+20)/ (4x+5)^2 = (-40)/ (4x+5)^2 6y = (-40)/(4x+5)^2 Spreading false math yo. Hmm, now stuck Link to comment Share on other sites More sharing options...
エメラルドK 696 Posted March 7, 2014 Share Posted March 7, 2014 if u need help with ur hw, then feel free to comment on this thread ) so yea i'll be the first one i need help with this problem: log2(y+4)+ log2y=5 how is it 4 and can som1 tell me how to determine if its extraneous or we You can use the addition/multiplication rule to make it log2(y2+4y)=5 then using log rules and making it an exponential function (we called it first to the last equals the middle) 25= y2+4y 0=y2+4y-32 Use sum product (so numbers would be 8 and -4). Idk what you call this but a type of factoring. 0= (x+8) (x-4) x=4 x=-8 (-8 is extraneous because when plugged back in it has no value) Link to comment Share on other sites More sharing options...
North and South 4,680 Posted March 7, 2014 Share Posted March 7, 2014 Oh I actually need some help as well. It's a basic basis physics problem, but we only get two tries on the assignment and I've used one lol Conductivity, a material's ability to conduct an electric current, is defined as the inverse of resistivity. Which of the following units represent conductivity?Enter all the items that are true, e.g., ABCD. A m2/( V m) Ω-1 cm-1 A m/( V cm2) kΩ-1 mm-1 ​I think the answer is BCD but tbh I'm not entirely sure. ​Some work I did, conductivity is in SI units of S/m where S= Ω-1 and S= A/V I always flunk the stupid questions. Any help would be really appreciated and i'd be happy to return the favour! Link to comment Share on other sites More sharing options...
dyke 270 Posted March 7, 2014 Author Share Posted March 7, 2014 Oh I actually need some help as well. It's a basic basis physics problem, but we only get two tries on the assignment and I've used one lol Conductivity, a material's ability to conduct an electric current, is defined as the inverse of resistivity. Which of the following units represent conductivity? Enter all the items that are true, e.g., ABCD. A m2/( V m) Ω-1 cm-1 A m/( V cm2) kΩ-1 mm-1 ​I think the answer is BCD but tbh I'm not entirely sure. ​Some work I did, conductivity is in SI units of S/m where S= Ω-1 and S= A/V I always flunk the stupid questions. Any help would be really appreciated and i'd be happy to return the favour! this is for algebra 2........ :/ Link to comment Share on other sites More sharing options...
Leo's Candy Jelly Ho 3,458 Posted March 7, 2014 Share Posted March 7, 2014 I sucked at math, EDIT: I HAVE REMEMBERED. *face palm* 2(3y)^(2-1) = (4x-5)(4) - (4x+5)(4)/ (4x+5)^2 = (16x-20) - (16x+20)/ (4x+5)^2 = (-40)/ (4x+5)^2 6y = (-40)/(4x+5)^2 Spreading false math yo. Hmm, now stuck LOL ummm I have no idea what you just did lmao Link to comment Share on other sites More sharing options...
Ghettoe 1,955 Posted March 7, 2014 Share Posted March 7, 2014 LOL ummm I have no idea what you just did lmao You have the quotient rule. The right side of the equation utilizes this rule because you have a fraction. The left side utilizes you standard derivative. You get 6y from bringing down the 2 in the 3y^2 and subtracting the 1. For the right side when you use the quotient rule, it is simpler to substitute values. For instance, have g(x) = 4x-5 and h(x) = (4x+5), then you plug it in. Basically, follow the formula. The derivative of g(x) times h(x) - g(x) times the derivative of h(x) divided by h(x) squared to put it in words. I actually think that is the full answer. I don't recall having to go further with the denominator. I haven't taken calculus in years though but I'm currently taking microeconomics with calculus and all we do is set up Lagrangians. Worse comes to worse, google the quotient rule and find similar problems online. Link to comment Share on other sites More sharing options...
Allegra 491 Posted March 7, 2014 Share Posted March 7, 2014 I sucked at math, EDIT: I HAVE REMEMBERED. *face palm* 2(3y)^(2-1) = (4x-5)(4) - (4x+5)(4)/ (4x+5)^2 = (16x-20) - (16x+20)/ (4x+5)^2 = (-40)/ (4x+5)^2 6y = (-40)/(4x+5)^2 Spreading false math yo. Hmm, now stuck You forgot the dy/dx on the left hand side because you have differentiated implicitly (so according to the chain rule you get dy/dx as well as the derivative of 6y^2...). Then rearrange to give dy/dx. If anyone needs maths/biol/chem/other help (even Latin), I'm happy to help as long as it's below university level as I'm in my last year of school. I can't help with analysis or anything, haha. Link to comment Share on other sites More sharing options...
Allegra 491 Posted March 8, 2014 Share Posted March 8, 2014 2(3y)^(2-1)*dy/dx= (4x-5)(4) - (4x+5)(4)/ (4x+5)^2 6y*dy/dx = (16x-20) - (16x+20)/ (4x+5)^2 = (-40)/ (4x+5)^2 6y*dy/dx = -40/(4x+5)^2 Rearrange... dy/dx = -40(4x+5)^2*6y Fixed. Once I've figured out how to use LaTex, life would be easier... Link to comment Share on other sites More sharing options...
dyke 270 Posted March 12, 2014 Author Share Posted March 12, 2014 ***BUMP***** For rational functions, what starter points should you use?????? Link to comment Share on other sites More sharing options...
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