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mizaki2100
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Post #570281 - Reply to ( #570279) by neonkitty |
The Weird Director
 Member
12:28 am, Sep 17 2012
Posts: 300
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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
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Post #570283 - Reply to ( #570281) by mizaki2100 |
Member
12:33 am, Sep 17 2012
Posts: 317
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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!  I really don't know why it's difficult for me to solve problems like this. xD
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WandereroftheDeep
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Post #570363 - Reply to ( #570279) by neonkitty |
Mmm...Tasty
Member
6:22 pm, Sep 17 2012
Posts: 485
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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?
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
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Post #570411 - Reply to ( #570363) by WandereroftheDeep |
Member
2:08 am, Sep 18 2012
Posts: 317
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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
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
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TheShawn
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Post #570414 - Reply to ( #570411) by neonkitty |
Member
2:21 am, Sep 18 2012
Posts: 247
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Quote from neonkitty Sweet, thanks a bunch 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
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Post #570423 - Reply to ( #570414) by TheShawn |
Member
2:41 am, Sep 18 2012
Posts: 317
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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 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
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TheShawn
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Post #570428 - Reply to ( #570423) by neonkitty |
Member
3:45 am, Sep 18 2012
Posts: 247
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Quote from neonkitty Thanks 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
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Post #570436 - Reply to ( #570428) by TheShawn |
Member
4:55 am, Sep 18 2012
Posts: 317
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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
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
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_baka_
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Post #570437 |
Member
5:11 am, Sep 18 2012
Posts: 50
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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
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Post #570444 |
the mu...
Member
7:01 am, Sep 18 2012
Posts: 973
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75%=180
100%=?
100%=(100/75)*180
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WandereroftheDeep
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Post #570448 |
Mmm...Tasty
Member
8:31 am, Sep 18 2012
Posts: 485
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@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
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Post #570536 - Reply to ( #570448) by WandereroftheDeep |
The Weird Director
Member
1:07 am, Sep 19 2012
Posts: 300
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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
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Post #570543 |
Member
1:53 am, Sep 19 2012
Posts: 47
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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. 
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chineserider
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Post #570546 - Reply to ( #570543) by crimsonsnow00017 |
Persona non grata
Member
2:35 am, Sep 19 2012
Posts: 1439
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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+1x->-2^- .......x+2 Without graphing.
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
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Post #570549 |
the mu...
Member
3:39 am, Sep 19 2012
Posts: 973
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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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