# Math Help - Caculus Homework Help!!

1. ## Caculus Homework Help!!

I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, 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

2. 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...

3. 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...

4. Originally Posted by Caycee7
I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, 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:
$\frac{x}{\sqrt{9x-10}}+2\sqrt{x-1}$

5. Originally Posted by Caycee7
I start my Ap Cal class tomarrow and our summer homework included some problems I just cannot figure out, 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
$
x(9x-10)^{-1/2} + 2(x-1)^{1/2}=\frac{x}{\sqrt{9x-10}}+ 2\sqrt{x-1}=
$
$\frac{x+2\sqrt{x-1}\sqrt{9x-10}}{\sqrt{9x-10}}
$

RonL

6. 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: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

7. 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

8. ## Got another one!

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!)

9. Originally Posted by Caycee7

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,
$x>-\frac{2}{3}$

10. 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?

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

12. 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...

13. 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?

14. 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)

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

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: $x\geq \frac{-3}{2}$

15. ## 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...