Quick maths questions Watch

MrToodles4
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Say you have sin(2x-30) =1 and you're trying to solve for x values within the range 0<x<180

If I want to change the range to first find the values what would I do first? I forgot if i first multiply 180*2 then -30... or do i subtract 30 first then multiple by 2.

I know once i have the x values eg. say 2x - 30 = 1/2Pi I would add 30 first then divide by 2 to get a solution for x - is changing the range the same?
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MrToodles4
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(Original post by Notnek)
Replace 0<x<180

by

0<2x-30<180

Then solve the inequality.
15<2x-30<105 ?
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dpksrk2015
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(Original post by MrToodles4)
Say you have sin(2x-30) =1 and you're trying to solve for x values within the range 0<x<180

If I want to change the range to first find the values what would I do first? I forgot if i first multiply 180*2 then -30... or do i subtract 30 first then multiple by 2.

I know once i have the x values eg. say 2x - 30 = 1/2Pi I would add 30 first then divide by 2 to get a solution for x - is changing the range the same?

yeah change the range so eventually you get 15<x<105
either draw the sine graph or use a cast diagram to work out your solutions but make them all equal to 2x-30 and then solve it. e.g. if one solution is idk like 20 then

2x-30= 20

therfore x = 25
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Notnek
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(Original post by MrToodles4)
15<2x-30<105 ?
Sorry I was a bit distracted and posted something very stupid

You need an inequality for 2x-30

If 0&lt;x&lt;180

Then multiplying by 2:

0&lt;2x&lt;360

Can you finish it?
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Notnek
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(Original post by dpksrk2015)
yeah change the range so eventually you get 15<x<105
This isn't right. I confused everyone with my post.
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RDKGames
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(Original post by MrToodles4)
Say you have sin(2x-30) =1 and you're trying to solve for x values within the range 0<x<180

If I want to change the range to first find the values what would I do first? I forgot if i first multiply 180*2 then -30... or do i subtract 30 first then multiple by 2.
Why not just try both and see which one gives you the expression you want? All you need to know is that whatever operation you do, you do it to the whole side.

This just means that if you subtract 30 to get -30<x-30<150 then multiplying through by 2 would yield 2(x-30) in the middle rather than 2x-30.
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MrToodles4
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(Original post by Notnek)
Sorry I was a bit distracted and posted something very stupid

You need an inequality for 2x-30

If 0&lt;x&lt;180

Then multiplying by 2:

0&lt;2x&lt;360

Can you finish it?
Its fine, I was thinking it seems to odd but the person below second' it so I was confused. So it should be -30<2x-30<330
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Notnek
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(Original post by MrToodles4)
Its fine, I was thinking it seems to odd but the person below second' it so I was confused. So it should be -30<2x-30<330
Yes that's right. Sorry again.
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MrToodles4
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(Original post by Notnek)
Yes that's right. Sorry again.
Its honestly fine. I really appreciate it thank you so much
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dpksrk2015
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(Original post by Notnek)
This isn't right. I confused everyone with my post.
I'm doing this in maths currently and I think it's okay? when you change the range you do it to both sides

0<2x-30<180

add 30 to both sides

30<2x<210

divide both sides by 2

15<x<105
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dpksrk2015
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(Original post by Notnek)
Yes that's right. Sorry again.
im so confused haha why don't you subtract first?? :albertein:
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Notnek
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(Original post by dpksrk2015)
im so confused haha why don't you subtract first?? :albertein:
If you start from 0<x<180 and subtract 30 you get

-30<x-30<150

Then multiply by 2

-60<2x-60<300


But that's an inequality for 2x-60 not 2x-30.
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MrToodles4
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(Original post by dpksrk2015)
im so confused haha why don't you subtract first?? :albertein:
I think its because if you subtract 30 first you get:

-30<x-30<150

and then to multiply by 2:

-60<2(x-30)<300

and so we got 2x-60 in the middle instead of 2x-30

So it makes sense to multiple by 2 first to get:

0<2x<360

and then take away 30

-30<2x-30<330
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RDKGames
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If you want to do subtraction before multiplication, then you need to subtract 15 from everything and only then mult by 2.
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