Need help w/ test for job: will take hours, so call, etc.

[Tedow]

I am in need of some major assistance ASAP. My employer has given all staff a take home "test" that is composed of mostly basic and intermediate algebra questions. I am in line for a promotion but am VERY weak at Algebra as it has been years since I have studied or used these concepts in school. I have attempted to learn these questions but I really have no base to begin and I truely believe it would take me year to fully get myself up to speed. I have been unable to find much help online and would even be willing to pay for assistance in these questions. I would anticipate that if someone would be willing to help me out...it should take a few hours. I can talk via phone, email, or instant messenger. One problem I have noticed already in trying to figure out these problems online is that it is rather difficult to type information in "mathmatical terms". Please email me at [deleted for security]

Ted

 

[Deleted member 4993]

Re: Please help ASAP

I am in need of some major assistance ASAP. My employer has given all staff a take home "test" that is composed of mostly basic and intermediate algebra questions. I am in line for a promotion but am VERY weak at Algebra as it has been years since I have studied or used these concepts in school. I have attempted to learn these questions but I really have no base to begin and I truely believe it would take me year to fully get myself up to speed. I have been unable to find much help online and would even be willing to pay for assistance in these questions. I would anticipate that if someone would be willing to help me out...it should take a few hours....

We are willing help you - by teaching you. However, we are not willing to help you by doing your problems and get you a promotion that will not be suitable for you.

Post your problems and your work (indicating exactly where you are stuck) - we will unstuck you and you will solve your problems.

 

[Tedow]

Re: Please help ASAP

My problem is I have NO idea how to type the problems Im working on to get the help....but here is what i have


simplify the expression

(4u^2v^2) -2
------------
(12u^-3v)


That should read something like this ...(everything in parenthesis except the negative 2 power) 4 u to the 2nd power v to the 2nd power divided by 12 u to the negative 3rd power v. Im not trying to get someone to do the problems for me..I just dont know how to type to get the help needed. I know that u is my base and that i know that i can simplify 4u and 12 u. But a little lost after that


Ted

 

[Deleted member 4993]

Re: Please help ASAP

My problem is I have NO idea how to type the problems Im working on to get the help....but here is what i have


simplify the expression

(4u^2v^2) -2
------------
(12u^-3v)


That should read something like this ...(everything in parenthesis except the negative 2 power) 4 u to the 2nd power v to the 2nd power divided by 12 u to the negative 3rd power v. Im not trying to get someone to do the problems for me..I just dont know how to type to get the help needed. I know that u is my base and that i know that i can simplify 4u and 12 u. But a little lost after that


Ted

I assume it looks like as follows:
\(\displaystyle [{\frac{4u^2v^2}{12u^{-3}v}}]^{-2}\)

if it then above is

\(\displaystyle = \, [{\frac{12u^{-3}v}{4u^2v^2}}]^{2}\)

\(\displaystyle = \, [{\frac{12}{4} \cdot \frac {u^{-3}}{u^2} \cdot \frac{v}{v^2}}]^{2}\)

continue from here...

 

[Loren]

Re: Please help ASAP

Laws of exponents that are involved in this problem...
(I) \(\displaystyle a^x\cdot a^y = a^{x+y}\)

(II) \(\displaystyle (a^x)^y = a^{xy}\)

(III) \(\displaystyle a^{-x} = \frac{1}{a^x}\), hence \(\displaystyle \frac{1}{a^{-x}} = a^x\)

\(\displaystyle \left(\frac{4u^2v^2}{12u^{-3}v}\right)^{-2}=\)

\(\displaystyle \left(\frac{1\cdot u^2\cdot u^3\cdot v^2\cdot v^{-1}}{3}\right)^{-2}}\) using "reduce fraction >> 4/12 to 1/3 and III.

\(\displaystyle \left(\frac{1\cdot u^{2+3}\cdot v^{2-1}}{3}\right)^{-2}}\) using I.

\(\displaystyle \left(\frac{u^5v^1}{3}\right)^{-2}\) Simplifying

\(\displaystyle \left(\frac{u^{-10}v^{-2}}{3^{-2}}\right)\) using II

\(\displaystyle \frac{3^2}{u^{10}v^2}\) using III

OR \(\displaystyle 3^2u^{-10}v^{-2}\) depending whether they ask for "all positive exponents" or "no fractions".

The order of applying the various laws of exponents makes no difference. Some people would prefer to get rid of the parenthesis and -2 exponent first. It's all a matter of preference. And of course, the 3^2 would probably be written as 9.

 

[skeeter]

Re: Please help ASAP

My employer has given all staff a take home "test" that is composed of mostly basic and intermediate algebra questions.

not making light of your situation, but I wish some of my students could foresee this very scenario occuring later in their adult lives.

 

[Denis]

Re: Please help ASAP

Dunno, but I gather that only the numerator is to the power -2; if so, then:
(4u^2v^2)^( -2) / (12u^(-3)v) .....you do realise I hope that 4u^2v^2 means 4 times u^2 times v^2 ??

= 1 / [(4u^2v^2)^2 * (12u^(-3)v)]

= 1 / [16 u^4 v^4 12 u^(-3) v]

= 1 / (192 u v^5)

To alleviate your typing uncertainty(!), Click on "Forum Help" at top, then on "Typing Math"

AND I'd like to add: thanks for telling us this may help you get a promotion; we seldom see such honesty here :idea:

 

[mmm4444bot]

Re: Please help ASAP

I dunno, Denis. The ambiguity of contemporary American English impedes.

... (everything in parenthesis except the negative 2 power) [expression] divided by [expression] ...

Ted, can you confirm the expression in Loren's post, please. Does it look correct? We need to know whether or not the entire fraction appears inside the parentheses because we need to know whether or not the entire fraction is being raised to the power of negative two OR if only the top of the fraction is being raised to the power of -2.

What line of work are you in?

Cheers,

~ Mark

 

[Denis]

Re: Please help ASAP

Well, really don't matter much at this point: we've shown Tedow (Ted ?) how to do it both ways.

We know it's in form (a / b)^(-2) OR a^(-2) / b

So Ted, "impress" your Boss with the following:

NOTE: Boss, that thing is not 100% clear; does the "^(-2)" apply to numerator only, or to the whole division?
If to numerator, then: [show work]
If to whole division, then: [show work]

CAUTION: he may fire you for accusing him of "unclarity"....

 

[Tedow]

Re: Please help ASAP

the entire question is in parenthesis except for the -^2 which is outside. I have not came up with the same answer as those that were provided. I ended up with:

(12u^-3v)^2 144u^-6v^3 9
________ ___________ = _____
(4u^2v2) = 16u^4v^4 u^10v^2


Obviously I screwed up majorily somewhere. Which of the above answers provided is correct for my particular question? I have a few others that I need help on which i will post soon. Thanks everyone for the help!

 

[mmm4444bot]

Re: Please help ASAP

the entire question is in parenthesis except for the -^2 which is outside.

Hi Tedow:

Okay -- thanks for clarifying that the entire fraction is being raised to the -2 power.

Two previous replies already explain how to deal with this exercise!

1) by Subhotosh Khan on Sun Aug 31, 2008 12:47 pm

2) by Loren on Sun Aug 31, 2008 1:36 pm

Do you understand these two replies? If NOT, then please be SPECIFIC.

Additionally, if you had used the [Preview] button before submitting your post, then you would have seen that your attempt to "draw" complex fractions on this site failed. We cannot use the [Space] key to "draw" stuff here because this web site strips out extra adjacent space characters.

Use the

tags to create a block of text where the extra spaces are not stripped out.

Or, use TEX:

[t e x]\frac{numerator goes here}{denominator goes here}[/ t e x]

Or, click on the Format Help drop-down menu and go to Kar'ls Notes to learn how to use grouping symbols {[()]} to type mathematical expressions without using web site coding features.

I'm not motivated to decipher your post from 6:24 p.m.

I would be motived to help you further if I knew why you don't understand the previous (noted) responses.

Cheers,

~ Mark

 

[Denis]

Re: Please help ASAP

Ted, you been drinking or something ? :wink:

Your answer of 9 / (u^10 v^2) is CORRECT, plus MATCHES the one Loren gave you.

Your equation typing is still horrific...PLEASE go to "Forum Help" to learn.