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mizaki2100
Post #570281 - Reply to (#570279) by neonkitty
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12:28 am, Sep 17 2012
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Quote from neonkitty
This years salary, $50,220, is an 8% increase over last year's salary. What was last year's salary?


$46,500

Eq I used:
50220 = x + .08x
50220 = 1.09x
Solving for x would get you 46500

Check:
46500 * .08 = 3720

3720 + 46500 = 50220



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neonkitty
Post #570283 - Reply to (#570281) by mizaki2100
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12:33 am, Sep 17 2012
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Quote from mizaki2100
46500

Eq I used:
50220 = x + .08x
50220 = 1.09x
Solving for x would get you 46500

Check:
46500 * .08 = 3720

3720 + 46500 = 50220

..Might be wrong too; it's been a while since I've done a math problem.


It's the right answer, thank you! biggrin I really don't know why it's difficult for me to solve problems like this. xD


WandereroftheDeep
Post #570363 - Reply to (#570279) by neonkitty
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6:22 pm, Sep 17 2012
Posts: 495


Quote from neonkitty
This years salary, $50,220, is an 8% increase over last year's salary. What was last year's salary?


I see you already have an answer, but this is an easier way to get there.

50,220/108 x100 = 46500

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neonkitty
Post #570411 - Reply to (#570363) by WandereroftheDeep
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2:08 am, Sep 18 2012
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Quote from WandereroftheDeep
I see you already have an answer, but this is an easier way to get there.

50,220/108 x100 = 46500



Sweet, thanks a bunch biggrin


Here's another horrendous math problem:
A= B/2+C/2 solve for B >_>"

I know the answer is B=2A-C but idk how to get that :l

TheShawn
Post #570414 - Reply to (#570411) by neonkitty
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2:21 am, Sep 18 2012
Posts: 247


Quote from neonkitty
Sweet, thanks a bunch biggrin


Here's another horrendous math problem:
A= B/2+C/2 solve for B >_>"

I know the answer is B=2A-C but idk how to get that :l


It's just a matter of setting the equation correctly. You're trying to solve for B, so the first step is to get the B variable by itself on one side of the equation.

A = B/2 + C/2

The easiest way is to subtract C/2 from both sides as if you were subtracting any whole number.

A - C/2 = B/2 + C/2 - C/2

That leaves us with:

A - C/2 = B/2

Now your goal is to solve for B, not B/2, so you multiply both sides of the equation by 2.

2 * (A - C/2) = 2 * (B/2)

Now we have:

2A - C = B

Which can also be written:

B = 2A - C.

And there we go. I mathed.



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neonkitty
Post #570423 - Reply to (#570414) by TheShawn
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2:41 am, Sep 18 2012
Posts: 318


Quote from TheShawn
It's just a matter of setting the equation correctly. You're trying to solve for B, so the first step is to get the B variable by itself on one side of the equation.

A = B/2 + C/2

The easiest way is to subtract C/2 from both sides as if you were subtracting any whole number.

A - C/2 = B/2 + C/2 ...


Thanks biggrin I found out how todo it a slightly different way, would this be wrong?
A=B/2+C/2
multiplied both sides by 2/1 so I got 2A=2B/2+2C/2
So I was left with 2A=B+C
Minus C by both sides and get
2A-C=B

TheShawn
Post #570428 - Reply to (#570423) by neonkitty
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3:45 am, Sep 18 2012
Posts: 247


Quote from neonkitty
Thanks biggrin I found out how todo it a slightly different way, would this be wrong?
A=B/2+C/2
multiplied both sides by 2/1 so I got 2A=2B/2+2C/2
So I was left with 2A=B+C
Minus C by both sides and get
2A-C=B


Yes, that would be fine. As long as you're applying the operation, such as multiplication in this case, to both sides of the equation, you can do it whenever you want throughout the process. Doing it at the beginning works just as well as at the end.

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neonkitty
Post #570436 - Reply to (#570428) by TheShawn
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4:55 am, Sep 18 2012
Posts: 318


Quote from TheShawn
Yes, that would be fine. As long as you're applying the operation, such as multiplication in this case, to both sides of the equation, you can do it whenever you want throughout the process. Doing it at the beginning works just as well as at the end.

Alright, thanks biggrin

Last question and I promise I'm done for the day xD

After a 25% reduction, you purchase a table for $180. What was the table's price before the reduction? Answer is 240 but I don't know how they got there. :l

_baka_
Post #570437
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5:11 am, Sep 18 2012
Posts: 50


probably a easier way to do this but given its a 25% reduction, 180 is the remaining 75% which can be written as 3/4 since this is also 75%, so you can divide by 3 to get its parts which is 60 and so 1/4 of the value is 60 which is 25% of the whole value, so 60X4 = 240

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mu2020
Post #570444
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7:01 am, Sep 18 2012
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75%=180
100%=?
100%=(100/75)*180

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WandereroftheDeep
Post #570448
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8:31 am, Sep 18 2012
Posts: 495


@neonkitty, you seem to be having problems with the same type of questions, so may I suggest a universal formula for dealing with percentages. It's actually really simple; you have a number x, which is a percentage y, of which you want to know a percentage z. If you put this into a formula you get this:
x / y . z = answer

With your most recent problem as an example that works out like this:

180 / 75 . 100 = 240

What mu2020 used is basically the same formula, but I thought this might be easier for you to understand.

Last edited by WandereroftheDeep at 8:38 am, Sep 18

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mizaki2100
Post #570536 - Reply to (#570448) by WandereroftheDeep
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1:07 am, Sep 19 2012
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Quote from WandereroftheDeep
you have a number x, which is a percentage y, of which you want to know a percentage z. If you put this into a formula you get this:
x / y . z = answer


After 3 years of not seeing math, and 5 years of not doing algebra stuff, I'm very thankful for reminding me the existence of this formula.

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crimsonsnow00017
Post #570543
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1:53 am, Sep 19 2012
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Find the limit, if it exists:

lim.....1-cos(x)
x->0.....sin(x)

and/or

from the left
lim...........2(x^2)+x+1
x->-2^- .......x+2

Without graphing. confused

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chineserider
Post #570546 - Reply to (#570543) by crimsonsnow00017
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2:35 am, Sep 19 2012
Posts: 1490


Quote from crimsonsnow00017
Find the limit, if it exists:

lim.....1-cos(x)
x->0.....sin(x)

and/or

from the left
lim...........2(x^2)+x+1
x->-2^- .......x+2

Without graphing. confused


For the first problem, use l'hopital's rule and take the derivative of the numerator over the derivative of the denominator. The answer to the second problem should be infinity because a real number over zero goes to infinity.

Not entirely sure of the answers though.

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mu2020
Post #570549
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3:39 am, Sep 19 2012
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1st problem
lim sin(x) = 0........................ if i'm not mistaken...................
x->0 cos(x)

as for the second one.... i think u mistyped the problem...

Last edited by mu2020 at 5:57 am, Sep 19

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