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I have not done this. I googled and it made sense soooo here it goes :P

6sqrtX=6x^(1/2)

Derivative being nx^n-1 so (1/2)6x^[(1/2)-1]

Giving you 3x^(-1/2) which is the same thing as 3/x^(-1/2) which is the same thing as 3/sqrtx

Derivative will give you the slope of the tangent which in this case would be 3/sqrt4=1.5

To create an equation you need some point on that line. You have x=4. Plug that in to the ORIGINAL equation to find your Y which would be 6sqrt4 or 12. So now you have (4, 12) as your point

y-12=1.5(x-4)

y-12=1.5x-6

y=1.5x+6

SUMMARY:

e0SBy.jpg

Edited by Drake Glau
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Need help with this, thanks:

Find the area bounded by the curves:

y= 2+x-x^2

y= x^2-3x+2

I just don't get the top-bottom thing.

:unsure:

I got something different.

aXMnc.gif

there you go nametaken :)

Bgr3m.jpg

Your method was right, your last line wasn't. I think you forgot to divide the 16 by two when you transformed the sixths into thirds.

Edited by Keel
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  • 2 weeks later...

Question:

In a game of Scrabble, Dalene has the seven letters A, D, E, k, O, Q and S. She picks two of these letters at random.

a) What is the probability that one is a vowel and the other is the letter D?

In my textbook it says the answer is 1/14. :blink:

Thanks.

P(D)=1/7

P(vowel after getting a D)=3/6

1/7 x 3/6 = 3/42 = 1/14

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  • 4 weeks later...

So I need to differentiate and I'm working on it, get an answer, and it's completely wrong. I don't know what i'm doing apparently but this is the only problem in this section that has given me troubles. Any help would be good :P

And I don't know how to upload the pretty math or whatever so here's a word doc instead >.<

Calc Help Upload.docx

Edited by Drake Glau
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Your first 2 steps are ok, but in the 3rd, better split it into 2 fractions and then it'l be easier to put the values from the nominator to denominator (after this operations the denominators of these 2 fractions rather won't be the same).

I know it seems confusing, but I'll try to paste the wroking here :)

post-19415-0-79051300-1302730920_thumb.j

Edited by Slovakov
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Yea, not what the book says XD I don't know where they are getting this.

I'm getting (x3 + 1 - 3x2x1/3)/[3x2/3(x3 +1)2]

If that all makes sense. How do you make your math pictures? I'm tired of all these brackets...

Book says my denominator is right, but it's also saying the numerator is 1-8x3 >.<

Edited by Drake Glau
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Say. . .I've got a quick question, Tangent to the curve y=x3(x-2)2 has the coordinates (-1,-9) which meets Normal to the same curve at (1,1) at the point P. Find the coordinates of P.

So, I just wanted to ask if you can either show me how step by step what needs to be done OR just give me the formula to do so.

Thank you very much.

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Find the equation of the tangent first. Find dy/dx when x=-1 (gradient of tangent).

Find the equation of the normal. Find dy/dx when x=1 (gradient of tangent). The gradient of normal is its negative reciprocal.

Solve both equations simultaneously to get coordinates of P.

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I give up. I will do this last minute. So then I can focus more. But I have managed to figure it out :)

So when dealing with equations like that that needs to use both either product rule or quotient rule, you use the chain rule first, then use the answer of that to apply the respective problem rule of it?

How do you differentiate (x-2)2????

Edited by Desy ♫
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