Anyone here good with math?

I just got that homework .doc file. It opens just fine in AbiWord
(hehe... who needs the overhead of OpenOffice -- just kidding)
Slackware 10 doesn't have OpenOffice, it has AbiWord, though)

I will start at the beginning.

The notes would be cool, too, if it doesn't put you out.

 

number one:
notice how when you move a factor from the numerator to the denominator, or vice-versa, the sign on the exponent changes...

 

This is how the second and third ones are rendered on my screen.

Obviously, I am missing something.

I will *** ASSUME *** that if no powers are given, as in the top one in this jpg, then the square root of the whole business is to be taken.

I will also assume, that if there is a small number in FRONT of the entire quantity, as in the bottom expression in the jpg in this post, that that quantity is to be taken to that root.

In any event, this seems to be ***very strange*** notation to me.
It must be that this .doc file is not displayed well except in some Microsoft program, or....

notation has changed since I practiced math....


.... or.... the teacher is very.....
......innovative

 

I had trouble posting that jpg image...

 

question on this one http://forums.majorgeeks.com/attachment.php?attachmentid=10753&stc=1


You put 5/3 + 2

then after that I see 5 + 6 / 3

the 11/3

Confused Wouldnt it be 7 / 3 ?? or is there another property im missing?

 

OK look at the third one, using everything I learned talking to you here

3sqrt(1880X^5Y^8)


Distribute the sqrt

3{(1080^1/2)(X^5+1/2)(Y^8+1/2)}

soo work that out and I get

3{(540)(X^3)(Y^4)}

Multiply the 3 I get


1620(3X^3)(3Y^4)

That right?

 

First, look at the number 1080. You can factor that number into factors, two of which are cubes. These cubes can be taken out of the cube root brackets....

Here is the factorization:

 

In this way, you take the "six cubed" out from under the cubed root sign (or the 1/3 power sign... same thing).

Then, you distribute the 1/3 power among all the other factors:

 

We add 5/3 to 2. In order to do that, we must get a common denominator between 5/3 and 2. That common denominator is 3

Here is another look at just the exponent there...

 

Jeepers!
Wookie, please read posts 53 and 54 in this thread, above...
I think I am NOT getting a good idea of these problems. This file is not being rendered correctly on my box, I guess.

I see NOTHING about a square root in that problem.
What software are you using to look at the HW1.doc file?

 

I had thought that we had agreed to treat the third one as if everthing that was in braces was taken to the third root:

 

good news, I rememberd I could never sell my math book from my first semester and that has all the basic stuff in it youll probably be hearing from me when I get to derivatives because the book we have doesnt explain crap its mostly just problems and like 200 pages if even.

Thanks for your help.

 

Ok I think I got most of my basics down. I cant figure out where to begin with the derivitives. I cant make anything out to relate to the first problem rom my notes, since all we do in class is copy whats on the board.

Find the derivitive of f(x) where f = 19

I think im supposed to pu tthat into the

f(x+(triangle)(x)-d(x) / (triangle)X
(you can see the problem on the worksheet.)

If you can help get me goin in that id appreciate it. THanks

 

The short answer to this is:
the derivative of f(x) where f=19
is 0.

If the function has no dependence on x, the function is a constant.
This means it doesn't change. You can put any value of x in, it doesn't matter, the function will ALWAYS have the value of 19.

So, the function doesn't change. So, the rate of change
(that is what the derivative is) is zero.

 

Wookie:
Notice that on the handout is has this:

constant rule d(a) = 0

Well, that applies in this case. here, a=19

 

WOW, this stuff makes my brain hurt.

 

No kidding!
Wait until he starts doing ODE's!

We will REALLY suffer then!

 

Yuck...I hate math.

ESPECIALLY algebra.

Algebra is icky.

 

Its one of the most useful subjects you can learn.

 

damnit I knew it was something I was overlooking. Thanks

 

OK theres no X so thats means its a constant cause theres no way for it to change, simple enough.

now if I look at 7x-4 I dont see a rule that looks anything like the syntax. but going off my notes im gonna come up with this.

7x follows the rule of d(x^n)

sooooo

7(1x^1-1) -0 the -0 because its a constant using the d(a) = 0

which would mean f^1 = 7x


am I completely wrong? I hope im right it will help my morale towards this

 

f(x) = 7x - 4
You are using the right rule, there, but...
d/dx {f(x)} = d/dx{7x - 4}

= d/dx{7x} - d/dx{4}

= 7 - 0

d/dx {f(x)} = 7

 

can you explain exactly what the g and f and a are?

like in a f)^1 = af^1



thanks

 

scratch that last one I figured out G is given in certain problems

 

I think I got this

ok f = 9x + 11

11 = 0

f = 9x

?

 

f(x) = 9x + 11
Take the derivative of both sides of the equation:

d/dx {f(x)} = d/dx {9x + 11}

for a sum, the derivative, "operator" if you will, will distribute:

d/dx {f(x)} = d/dx {9x} + d/dx {11}

the left hand side of the equation is the derivative of f(x) with respect to x. No bones about it.

The right hand side...

d/dx {9x} = 9
this is from:

simple power rule d (xn) = n xn-1 dx

the last term on the right hand side:

d/dx {11} = 0
because the derivative of a constant is zero.
this comes from this rule:

constant rule d(a) = 0

 

f(x) = 9x + 11
Take the derivative of both sides of the equation:

d/dx {f(x)} = d/dx {9x + 11}

for a sum, the derivative, "operator" if you will, will distribute:

d/dx {f(x)} = d/dx {9x} + d/dx {11}

the left hand side of the equation is the derivative of f(x) with respect to x. No bones about it.

The right hand side...

d/dx {9x} = 9
this is from:

simple power rule d (xn) = n xn-1 dx

the last term on the right hand side:

d/dx {11} = 0
because the derivative of a constant is zero.
this comes from this rule:

constant rule d(a) = 0

so, in the end, you have this;

d/dx {f(x)} = 9