Caculus Homework Help!!

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August 6th 2006, 12:36 PM
Caycee7

Caculus Homework Help!!

I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, :confused: any help would be amazing!

~factor and simplify. express answer as a fraction withought neg. exponiets
x(9x-10)^-1/2 + 2(x-1)^1/2

~Another one is this... I thnk you may have to use compleating the squares but couldnt figure it out)

x^2-x=5
August 6th 2006, 12:43 PM
Quick

Quote:

Originally Posted by Caycee7

~Another one is this... I thnk you may have to use compleating the squares but couldnt figure it out)

x^2-x=5

subtract 5 from both sides: $x^2-x-5=0$

now use the quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

substitute: $x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-5)}}{2(1)}$

simplify: $x=\frac{1\pm\sqrt{1+20}}{2}$

you can solve from there...
August 6th 2006, 12:47 PM
Caycee7

Thanks so much for the help
but I dont really understand why you use the quadratic, :(
I havnt seen that formula in over a year and dont know how it relates...
August 6th 2006, 12:51 PM
Quick

Quote:

Originally Posted by Caycee7

I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, :confused: any help would be amazing!

~factor and simplify. express answer as a fraction withought neg. exponiets
x(9x-10)^-1/2 + 2(x-1)^1/2

first thing you need to remember is $(x^a)^b=x^{(a\times b)}$

second thing you need to remember is $x^{\frac{1}{2}}=\sqrt{x}$

therefore: $(9x-10)^{-\frac{1}{2}}=\left((9x-10)^{\frac{1}{2}}\right)^{-1}$ $=(\sqrt{9x-10})^{-1}=\frac{1}{\sqrt{9x-10}}$

therefore you get:
$\frac{x}{\sqrt{9x-10}}+2\sqrt{x-1}$

and solve from there, (if you can't just ask me to help you further)
August 6th 2006, 12:52 PM
CaptainBlack

Quote:

Originally Posted by Caycee7

I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, :confused: any help would be amazing!

~factor and simplify. express answer as a fraction withought neg. exponiets
x(9x-10)^-1/2 + 2(x-1)^1/2

$<br /> x(9x-10)^{-1/2} + 2(x-1)^{1/2}=\frac{x}{\sqrt{9x-10}}+ 2\sqrt{x-1}=<br />$ $\frac{x+2\sqrt{x-1}\sqrt{9x-10}}{\sqrt{9x-10}}<br />$

RonL
August 6th 2006, 12:53 PM
Quick

Quote:

Originally Posted by Caycee7

Thanks so much for the help
but I dont really understand why you use the quadratic, :(
I havnt seen that formula in over a year and dont know how it relates...

when given a quadratic equation: $ax^2+bx+c=0$

then you can use the quadratic formula to find the value(s) of x: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
August 6th 2006, 01:03 PM
Caycee7

Quote:

Originally Posted by Quick

when given a quadratic equation: $ax^2+bx+c=0$

then you can use the quadratic equation to find the value(s) of x: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Oh okay that makes sense, wow it is amazing how rusty you get over the summer! Thanks so much you guys! I might have a few more in a bit, I have a few more pages to go hopefully I will be able to figure them out but if not you'll be hearing from me again ;)
August 6th 2006, 01:21 PM
Caycee7

Got another one!

the question reads:

find the domain of the function

f(x)=ln(3x+2)

(Ahh I dont remember what find the domain means... Am I suppose to solve for x instead of f(x)... That is all I recall domain being, x on a graph!)
August 6th 2006, 01:23 PM
ThePerfectHacker

Quote:

Originally Posted by Caycee7

the question reads:

find the domain of the function

f(x)=ln(3x+2)

(Ahh I dont remember what find the domain means... Am I suppose to solve for x instead of f(x)... That is all I recall domain being, x on a graph!)

Domain means what values of "x" can take, basically.

Thus, since $\ln (3x+2)$ can only make sense when the expression inside the parantheses is positive. Thus,
$3x+2>0$
Thus,
$x>-\frac{2}{3}$
August 6th 2006, 01:43 PM
Caycee7

Okay I feel bad to keep posting questions but I have one that is asking me for the general form of a line passing through pint (3,1) slope=5/2

I thought this isnt bad and drew the graph but now I found out that that is a formula that I have never studied at all, I tried to read up on it on some other sites but got extreamly confused, and ideas?
August 6th 2006, 01:47 PM
Quick

Quote:

Originally Posted by Caycee7

Okay I feel bad to keep posting questions but I have one that is asking me for the general form of a line passing through pint (3,1) slope=5/2

I thought this isnt bad and drew the graph but now I found out that that is a formula that I have never studied at all, I tried to read up on it on some other sites but got extreamly confused, and ideas?

label your point as $(x_1,y_1)$ then make an equation in point-slope form: $y-y_1=m(x-x_1)$

now substitute: $y-1=\frac{5}{2}(x-3)$
August 6th 2006, 01:56 PM
Caycee7

Quote:

Originally Posted by ThePerfectHacker

Domain means what values of "x" can take, basically.

Thus, since $\ln (3x+2)$ can only make sense when the expression inside the parantheses is positive. Thus,
$3x+2>0$
Thus,
$x>-\frac{2}{3}$

So if that is true the domain of f(x)= the square root of 2x+3

does this mean x> or equal to -4?

I really need to figure out the math typing thing but you know what I mean...
August 6th 2006, 01:59 PM
Caycee7

Quote:

Originally Posted by Quick

label your point as $(x_1,y_1)$ then make an equation in point-slope form: $y-y_1=m(x-x_1)$

now substitute: $y-1=\frac{5}{2}(x-3)$

So the point slope form is simply the gen. form of the line?
August 6th 2006, 02:06 PM
Quick

Quote:

Originally Posted by Caycee7

So the point slope form is simply the gen. form of the line?

it's a general form of a line, I don't think there is one specified "general form" (although y=mx+b would be most likely)

---------------------------------------

to answer your other question:
you are given the equation $f(x)=\sqrt{2x+3}$

therefore: $2x+3\geq0$

subtract 3 from both sides: $2x\geq -3$

divide 2 from both sides: $x\geq \frac{-3}{2}$
August 6th 2006, 02:32 PM
Caycee7

Ahh Relief!

Thanks everyone I finally finished all of my problems! Took a few hours but a relief to have it all finished. Now I am off to finish my English work, I have a feeling it is going to be a long night! Oh well that’s what I get for procrastinating. Thanks again, I'm sure I'll be around now that I found this site, I have AP both for Cal and Stats, and math isn’t my best subject...

:D

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