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Lexyvil

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I got a Math test coming up very soon, and I'm left with a review sheet for practice. Sadly, I hardly know the procedures on how to solve them...

 

It involves isolating variables, logarithms and isolating from under a square root.

 

1. b- http://prntscr.com/7xto4e

 

1. c- http://prntscr.com/7xto9k

 

2. a- http://prntscr.com/7xtocq (This one, the "ln" and the minus "3" in front of it is what I don't get...)

 

2. b- http://prntscr.com/7xton4(How can this one be solvable anyway?)

 

2. c- http://prntscr.com/7xtotf(As I keep moving from page to page, I get scared from my inability of dealing with isolating variables within denomenators)

 

2. d- http://prntscr.com/7xtp3m(I think I can do this one since "e" is simply a constant).

 

And my biggest fear so far: http://prntscr.com/7xtyof (Whaaa..?)

 

The protocols on solving them is what boggles me. I do have the answer sheet, but it's the procedures on solving them that gets me kind of scratching me head. Any help on any of the questions or some, on how to solve them will help x   x;

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Guest -Shizuku-

Ugh... those are quite simple things although I would need to go back 5 years back, when I was in middle school to remember all this. If anyone wont lend you a hand you can just ask me for skype later. Denomenators - I know how to do this but to put it in english will be hardest lol... if you'd give me time then sure no problem.

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I studied law just so I could avoid this type of shit :P good luck with your test!

If i'm not mistaken this are pretty simple thou, I still don't know who to solve them xD but there should be one user in this forum that can help you out, I don't remember the user but there was someone here who knew math...it was a mod I think.

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I got a Math test coming up very soon, and I'm left with a review sheet for practice. Sadly, I hardly know the procedures on how to solve them...

 

It involves isolating variables, logarithms and isolating from under a square root.

 

1. b- http://prntscr.com/7xto4e

 

1. c- http://prntscr.com/7xto9k

 

2. a- http://prntscr.com/7xtocq (This one, the "ln" and the minus "3" in front of it is what I don't get...)

 

2. b- http://prntscr.com/7xton4(How can this one be solvable anyway?)

 

2. c- http://prntscr.com/7xtotf(As I keep moving from page to page, I get scared from my inability of dealing with isolating variables within denomenators)

 

2. d- http://prntscr.com/7xtp3m(I think I can do this one since "e" is simply a constant).

 

And my biggest fear so far: http://prntscr.com/7xtyof (Whaaa..?)

 

The protocols on solving them is what boggles me. I do have the answer sheet, but it's the procedures on solving them that gets me kind of scratching me head. Any help on any of the questions or some, on how to solve them will help x   x;

1b) Step one is to make everything linear, and equal 0:

Cross multiply the bottoms, netting you (2y-5)(8-2y)=(-3y+1)(y-4),

Multiply out. [16y-40-(2y2)+10y=(-3y2)+12y+y-4]=[-2y2+26y-40=-3y2+13y-4]

Move everything to one side by subtracting the right: y2+13y-36=0

Step two: Realize you have a quadratic in the form of ax2+bx+c=0

Use quadratic formula: x=-b +-((b2-4ac)/2a).5

At this point you use a calculator to solve. Alternatively, if you don't have a calculator, you can complete the square.

 

1c) I really can't remember.  You probably have to do something along the lines of using the difference of two squares.  Honestly, if this came up on a test, I would try plugging in every number from -5 to 5 in an attempt to save time rather than solving it manually.  Due to the nature of the questions on this sheet, and all the answers being whole numbers, I would guess this one is too, and as it turns out, it is, with the answer being 4.

 

2a) The ln is the natural logarithm, and the 3 is part of a logarithm's identities.

 

2b)They probably want you to solve for x, so square both sides: y2=4-2x

and then solve for x: x= .5y2-2

 

2c)again, solve for x

subtract 2 from both sides: y-2=1/(x+5)

reciprocate both sides, or alternatively divide 1 by both sides: 1/(y-2)=x+5

subtract 5: x=1/(y-2)-5

 

2d) again, another logarithmic identity, the third one, then the first.  This is because ln(x) is loge(x)

 

Final:

This one is really easy and requires no math at all, these are simply nested equations:

(a)f-1 is simply the reciprocal of the equation.  Because f(2)=2, f-1(2)=1/2

 

(b)this is simply solved by doing one step at a time working your way outwards:

f(  [  f(0) = -2  ]  ) = f(-2) = 1

 

© apply both a and b

f(  [  f-1(-1) = 1/0 = und  ]  ) = f(und) = und

 

(d) same as c

f-1(  [  f(1)=-1 ] ) = f-1(-1) = 1/-1 = -1

 

Hope this helps, and good luck on your test.  When working on your worksheet, if you ever come across something you don't know how to do, you can type the equation into Wolfram Alpha, solve, and one of the options under the answer is "Step by step Solution".  Click it, and wolfram will give you the first step, and the text portion of the second step which should give you an idea of what to do next.  You might be able to repeatedly do this over and over to find the solution.

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My solutions for the first two problems.

Looks like the other ones have been answered well enough.

 

1b) http://i.imgur.com/RLHUBNA.png

1c) http://i.imgur.com/4W3FPF1.png

 

 

EDIT:

 

Also, if problem 7 is using the inverse function for f-1 (and not reciprocating), then

7a)  f-1(2) = 2

7b) f(f(0)) = f(-2) = 1

7c) Nesting a function with its inverse will cancel them out and give the argument: it's just -1

7d) Likewise with 7c, it's just 1.

Edited by binaryfail
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Seems to me that the first parts were solved already. Also, I can't see the last part. It brings me back to the website upload page.

 

As for the second one. Graph transformation is done based on the general form of the equation. If I'm not mistaken, the things you have to change with the general form is usually the vertical shift, horizontal shift, and the multiplier. I think the transitional graph means that you have to apply these changes one at a time.

As for the domain and range, domain IIRC refers to x, so solve for y and determine the values possible for x. For the range, do the inverse. Solve for y and check for the values possible for x. Usually, we check the values that won't create an imaginary result and infinity. For the intercepts, to get the x-intercept, let y = 0 and for the y-intercept, let x = 0. To get the asymptotes, check for the values that will make y infinity.

 

I took this course way back in my BS days, so feel free to correct me if I made some mistakes. Feel free to PM me as well if you're having a hard time about my explanations.

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  • 3 weeks later...

An apparent rollback happened (I think?) so I'll say it again here: I was extremely busy with college so I couldn't really have the time to thank you all for the help, I did read some of what you all said and I did grasp a few new concepts~

 

I passed this excruciating summer semester so that's taken care of ^ ^

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