Does anybody know how you would make 'k' the subject of the equation? See the pic attached. Any help would be appreciated..
http://i51.tinypic.com/2mnj3uf.jpg
http://i51.tinypic.com/2mnj3uf.jpg
I know how but it would be better if you did your own homework.
Remember what you do to one side of the = sign you do to the other side.
hint....multiply both sides by k then divide each side by "a"
k=????????/
Remember what you do to one side of the = sign you do to the other side.
hint....multiply both sides by k then divide each side by "a"
k=????????/
k=sqr(b)/a
multiply both sides by root b then turn the whole thing upside down..
�� Quote: Originally Posted by Copyright 
k=sqr(b)/a
You get an "A" but go to the back of the room...Let boomstick figure it out!
�� Quote: Originally Posted by The Howling Wolves I know how but it would be better if you did your own homework.
Remember what you do to one side of the = sign you do to the other side.
hint....multiply both sides by k then divide each side by "a"
k=????????/
Remember what you do to one side of the = sign you do to the other side.
hint....multiply both sides by k then divide each side by "a"
k=????????/
I need an example so can do the rest...Does the sqr'b' stay the same even though it swapped sides? I thought it would become b^2?
Does making k the subject of the equation mean the same as solving for k? Is this the new math?
Find out what you each is asking for then do whatever it takes to get it alone on one side.
Then do the same to the other side..
In the first one you multiplied both sides by k
then divided by "a" to both sides to get k=b^2/a
Do that to the rest of them....
Then do the same to the other side..
In the first one you multiplied both sides by k
then divided by "a" to both sides to get k=b^2/a
Do that to the rest of them....
�� Quote: Originally Posted by The Howling Wolves Find out what you each is asking for then do whatever it takes to get it alone on one side.
Then do the same to the other side..
In the first one you multiplied both sides by k
then divided by "a" to both sides to get k=b^2/a
Do that to the rest of them....
Then do the same to the other side..
In the first one you multiplied both sides by k
then divided by "a" to both sides to get k=b^2/a
Do that to the rest of them....
the answer to Q.a is the easiest, since it just involves shifting the three expressions around
Q.b is slightly more complex since it involves balancing the equation by doing some arithmetic
- remember that what you do to one side of the equation, you must also do to the other..
so - for example, taking 2 from the right side means that you also have to do the same to the left..
you are not being asked to simplify the equations - just to rearrange them..
Q.b is slightly more complex since it involves balancing the equation by doing some arithmetic
- remember that what you do to one side of the equation, you must also do to the other..
so - for example, taking 2 from the right side means that you also have to do the same to the left..
you are not being asked to simplify the equations - just to rearrange them..
Howling Wolf is right...Basic rule with equations is whatever you do...always do it to both sides
@sportscout: apparently that's what they call it now. when I was at school, it was solve for k...For some reason "they" think they have to change terminology every decade (or so).
@Bugmeiser: Close.. but you might want to rethink that procedur
Edit, since you postet again: I refer to this
-DG
@sportscout: apparently that's what they call it now. when I was at school, it was solve for k...For some reason "they" think they have to change terminology every decade (or so).
@Bugmeiser: Close.. but you might want to rethink that procedur
Edit, since you postet again: I refer to this
Quote:
multiply both sides by root b then turn the whole thing upside down
Im still lost lol, if someone could break down the steps it would be awesome. Once I have the method figured out im sure these will seem easy, its all new to me I have no idea how to do a or b.. I know about balancing the sides, 'the basics' I don't understand how in question 'a' how the Square root of b stays as the square root of b if it changes sides etc..
�� Quote: Originally Posted by SledgeDG Howling Wolf is right...Basic rule with equations is whatever you do...always do it to both sides
@sportscout: apparently that's what they call it now. when I was at school, it was solve for k...For some reason "they" think they have to change terminology every decade (or so).
@Bugmeiser: Close.. but you might want to rethink that procedur
Edit, since you postet again: I refer to this
-DG
@sportscout: apparently that's what they call it now. when I was at school, it was solve for k...For some reason "they" think they have to change terminology every decade (or so).
@Bugmeiser: Close.. but you might want to rethink that procedur
Edit, since you postet again: I refer to this
Quote:
multiply both sides by root b then turn the whole thing upside down
With an operation involving roots you always have 2 options:
Option 1 : solve the root (extremely unpracticable)
Option 2: use the whole thing as 1 expression (which means it doesn't change when its been "moved" to the other side
I hope that's a bit clearer
-DG
Option 1 : solve the root (extremely unpracticable)
Option 2: use the whole thing as 1 expression (which means it doesn't change when its been "moved" to the other side
I hope that's a bit clearer
-DG
�� Quote: Originally Posted by Boomstick777 Im still lost lol, if someone could break down the steps it would be awesome. Once I have the method figured out im sure these will seem easy, its all new to me I have no idea how to do a or b.. I know about balancing the sides, 'the basics' I don't understand how in question 'a' how the Square root of b stays as the square root of b if it changes sides etc..
- it's a bit like the Time = Distance/Speed formula..
all you're doing is moving the expressions around to suit their use:
Distance = Speed.Time
Speed = Distance/Time
�� Quote: Originally Posted by Boomstick777 Im still lost lol, if someone could break down the steps it would be awesome. Once I have the method figured out im sure these will seem easy, its all new to me I have no idea how to do a or b.. I know about balancing the sides, 'the basics' I don't understand how in question 'a' how the Square root of b stays as the square root of b if it changes sides etc..
We broke down the first one now you have to do the rest...
If I fish for you ............you won't learn how to fish ...you will starve!
Ok I think I got question 'a' figured out..
a= Sqr(b)/k
becomes
a x k = Sqr(b) xk
a x k = Sqr(b)
a x k/a = Sqr(b)/a
so
k = Sqr(b)/a
Question 'b' is harder how do I just get the 'b' from the sqr(b)? Hmm
a= Sqr(b)/k
becomes
a x k = Sqr(b) xk
a x k = Sqr(b)
a x k/a = Sqr(b)/a
so
k = Sqr(b)/a
Question 'b' is harder how do I just get the 'b' from the sqr(b)? Hmm
I'd just leave it as root b
a-2= root of b over k
k(a-2) = root of b
k(a-2) = root of b
Remember: sqr(b)*sqr(b) = b
�� Quote: Originally Posted by SledgeDG Remember: sqr(b)*sqr(b) = b
I am amazed that I can still recall this being I graduated from HS in 1962!
College in 71 with serving our country in between.
�� Quote: Originally Posted by Boomstick777 ...Question 'b' is harder how do I just get the 'b' from the sqr(b)? Hmm
If you had sqr(9) on one side, what then?
Square both sides. Your sqr(9) magically becomes just "9", because:
sqr(9) X sqr(9) = 9
Remember what BugMeister said:
Quote:
...what you do to one side of the equation, you must also do to the other...
EDIT:
�� Quote: Originally Posted by SledgeDG Remember: sqr(b)*sqr(b) = b
Umm..... I'm not a mathemetician, sorry :/
- on reflection, neither am I
heh, heh..!!
heh, heh..!!
@THW:I didn't want to give him the result...rather have him recall basic rules. sorry if in doing so I spoiled your plan.
I'm actually enjoying this...haven't had the need to recall those calculations for pretty much an eternity now either
@Prof: my explanation is shorter... true but some ppl might understand better if they go the long way around
Perhaps I should register with Big Y (they run a homework helpline over here)
-DG
I'm actually enjoying this...haven't had the need to recall those calculations for pretty much an eternity now either
@Prof: my explanation is shorter... true but some ppl might understand better if they go the long way around
Perhaps I should register with Big Y (they run a homework helpline over here)
-DG
�� Quote: Originally Posted by BugMeister - on reflection, neither am I
heh, heh..!!
heh, heh..!!
�� Quote: Originally Posted by Benjamin Hall Umm..... I'm not a mathemetician, sorry :/
(Easier said than done, I know.)
It's no fun, but LOTS of practice is the best idea. You wouldn't shoot one free throw, sink it, then decide you had it down pat. But doing it over and over and over and...
Don't confuse me with your foreign language talk D=
Ehhhh
Ehhhh
What? Don't they play basketball in Wisconsin? 
No, what he's saying is in Wisconsin, people don't need to practice at anything. We always get it done right the first time.
I miss math so much I feel like doing your homework LOL. Let me think about it. Hmmm..
EDIT: So I wanted to use MS Math so I can template equations so it's easier to understand and that solve it a bit, but that thing freaking did it for me. So here it is. I'm going to leave the last 2 steps out since I don't want to spoon feed. The last number is up to you okay?
EDIT: So I wanted to use MS Math so I can template equations so it's easier to understand and that solve it a bit, but that thing freaking did it for me. So here it is. I'm going to leave the last 2 steps out since I don't want to spoon feed. The last number is up to you okay?
Is this the correct answer for question 'b'
b = k/a + 2
I used this method
a = sqr(b)/k + 2
a x sqr(b) = sqr(b)/k + 2 xsqr(b)
a x b = k + 2
a x b/a = k/a +2
Answer - b = k/a +2
b = k/a + 2
I used this method
a = sqr(b)/k + 2
a x sqr(b) = sqr(b)/k + 2 xsqr(b)
a x b = k + 2
a x b/a = k/a +2
Answer - b = k/a +2
When did they add letters
Nah. I suck at Algebra.
Let's go through this step by step.
a = sqr (b) / k + 2
sqr (b) / k = a - 2
sqr (b) = k (a -2)
sqr (b) = ka - 2k
b = (ka - 2k) x (ka - 2k)
b = k�a� - 4k�a + 4k� (answer)
Check using a = 6 and k = 3
b = (3� x 6�) - (4 x 3� x 6) + (4 x 3�)
b = 324 - 216 + 36
b = 144
Substituting b = 144 into the first equation, a = sqr (b) / k + 2, we get
a = (sqr (144) / 3) + 2
a = (12 / 3) + 2
a = 4 + 2
a = 6
So the answer is
b = k�a� - 4k�a + 4k�
Not bad, considering that it is about 30yrs since I did any Algebra. Hopefully, this should help you with the rest of the questions if you approach them in a similar way. Don't forget to do the checking, substituting actual numbers, as shown above. You will need to work out for yourself suitable numbers to use at this step.
a = sqr (b) / k + 2
sqr (b) / k = a - 2
sqr (b) = k (a -2)
sqr (b) = ka - 2k
b = (ka - 2k) x (ka - 2k)
b = k�a� - 4k�a + 4k� (answer)
Check using a = 6 and k = 3
b = (3� x 6�) - (4 x 3� x 6) + (4 x 3�)
b = 324 - 216 + 36
b = 144
Substituting b = 144 into the first equation, a = sqr (b) / k + 2, we get
a = (sqr (144) / 3) + 2
a = (12 / 3) + 2
a = 4 + 2
a = 6
So the answer is
b = k�a� - 4k�a + 4k�
Not bad, considering that it is about 30yrs since I did any Algebra. Hopefully, this should help you with the rest of the questions if you approach them in a similar way. Don't forget to do the checking, substituting actual numbers, as shown above. You will need to work out for yourself suitable numbers to use at this step.
Nice Job Dwarf!!
anybody good at washing up?
i've got a nice pile of dirty dishes growing in my kitchen..
i've got a nice pile of dirty dishes growing in my kitchen..
�� Quote: Originally Posted by Kirsch Nice Job Dwarf!!
�� Quote: Originally Posted by mickey megabyte anybody good at washing up?
i've got a nice pile of dirty dishes growing in my kitchen..
i've got a nice pile of dirty dishes growing in my kitchen..
�� Quote: Originally Posted by Dwarf Let's go through this step by step.
a = sqr (b) / k + 2
sqr (b) / k = a - 2
sqr (b) = k (a -2)
sqr (b) = ka - 2k
b = (ka - 2k) x (ka - 2k)
b = k�a� - 4k�a + 4k� (answer)
Check using a = 6 and k = 3
b = (3� x 6�) - (4 x 3� x 6) + (4 x 3�)
b = 324 - 216 + 36
b = 144
Substituting b = 144 into the first equation, a = sqr (b) / k + 2, we get
a = (sqr (144) / 3) + 2
a = (12 / 3) + 2
a = 4 + 2
a = 6
So the answer is
b = k�a� - 4k�a + 4k�
Not bad, considering that it is about 30yrs since I did any Algebra. Hopefully, this should help you with the rest of the questions if you approach them in a similar way. Don't forget to do the checking, substituting actual numbers, as shown above. You will need to work out for yourself suitable numbers to use at this step.
a = sqr (b) / k + 2
sqr (b) / k = a - 2
sqr (b) = k (a -2)
sqr (b) = ka - 2k
b = (ka - 2k) x (ka - 2k)
b = k�a� - 4k�a + 4k� (answer)
Check using a = 6 and k = 3
b = (3� x 6�) - (4 x 3� x 6) + (4 x 3�)
b = 324 - 216 + 36
b = 144
Substituting b = 144 into the first equation, a = sqr (b) / k + 2, we get
a = (sqr (144) / 3) + 2
a = (12 / 3) + 2
a = 4 + 2
a = 6
So the answer is
b = k�a� - 4k�a + 4k�
Not bad, considering that it is about 30yrs since I did any Algebra. Hopefully, this should help you with the rest of the questions if you approach them in a similar way. Don't forget to do the checking, substituting actual numbers, as shown above. You will need to work out for yourself suitable numbers to use at this step.
Nice job going step by step in solving his problem.
Could have done that early on in the post but thought it was important for the OP to practice and learn on his own. Isn't that the way we learned how to algebra?
I'm leaving the rest for him to work out on his own. Besides, the image isn't very clear with the other equations.
�� Quote: Originally Posted by The Howling Wolves ...thought it was important for the OP to practice and learn on his own. Isn't that the way we learned how to algebra?
Much easier.
(I made that up, Benjamin. Don't get any ideas.)
Was looking for something else and ran across this little sign...may have to click on it!
Enjoy your day...
Enjoy your day...
Just noticed that I can reduce the final answer down a bit further. I'll leave him to work that out for himself.
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