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cassi999 · Oct 13, 2011, 02:37 PM

transpose formula maths
transpose the formula to make z the subject and determine it value when w=4, x=1.5 and y=0.85

y=(2 x)(3-z)/4w

transpose the formula using z the subject and obtain its value when y=2*10^3 and x=750

y=3?(x^2 -z^2)

transpose the equations to make c the subject and determine its value when a=4.5 and b3.6
for a c^ =c^ 3 a b
b) transpose the equation to make v the subject and determine its value when a=0.75,u=3.6 and t=10.5

v^-u^/2a= u t 1/2 a t^

Unknown008 · Oct 14, 2011, 01:33 AM

1. To make z the subject of formula,
a. multiply both sides by 4w
b. divide both side by (2+x)
c. subtract 3 from both sides
d. multiply both sides by -1.

Can you try that an post what you get? :)

cassi999 · Oct 15, 2011, 02:25 PM

Really tired got random and weird answers.. could you try it.. and there was a mistake in the questions above .

a.) y=(2+x)(3-z)/4w
b.) y=3(x^2 -z^2)

Thank you

Unknown008 · Oct 16, 2011, 06:52 AM

It's just as I said.

Here, let me show you, but promise me to try the others and post every line you come up with :)

$y=\frac{(2+x)(3-z)}{4w}$

1. To make z the subject of formula,
a. multiply both sides by 4w

$4wy = \frac{(2+x)(3-z)}{\cancel{4w}}\times \cancel{4w}$

$4wy =(2+x)(3-z)$

b. divide both side by (2+x)

$\frac{4wy}{(2+x)} = \frac{\cancel{(2+x)}(3-z)}{\cancel{(2+x)}}$

$\frac{4wy}{(2+x)} =3-z$

c. subtract 3 from both sides

$\frac{4wy}{(2+x)} -3 =\cancel{3}-z \cancel{-3}$

$\frac{4wy}{(2+x)} -3=-z$

d. multiply both sides by -1.

$-1\times (\frac{4wy}{(2+x)} -3)=-1(-z)$

$-\frac{4wy}{(2+x)} +3=z$

$z = 3 -\frac{4wy}{(2+x)}$

Now, if you understood why I did those, you would see that each step removed something that was 'attached' to z. Can you try out the others and post them here? Even the weird answers that you got, and how you got them, so that we know exactly what went wrong.

cassi999 · Oct 16, 2011, 03:15 PM

okay.. I don't know how to convert steps into the formula with this equation... y=3(x^2 -z^2)

y=3(x 2 -z 2 )
(x 2 - z 2 ) = y/3
z =((x 2 - (y/3))
so that's (750^2 -z^2)-2*10^3/3
... 750^2-2000/3= 56183

I know something is wrong with the answer.. it looks funny

Unknown008 · Oct 16, 2011, 11:04 PM

Originally Posted by cassi999

okay.. I don't know how to convert steps into the formula with this equation... y=3(x^2 -z^2)

y=3(x^2 -z^2)
(x 2 - z 2 ) = y/3

Good :)

z =((x 2 - (y/3))

No, you missed the square here, or you forgot to type it... but anyway, this should be:

$z^2 = x^2 - \frac{y}{3}$

$z= \sqrt{$x^2 - \frac{y}{3}$}$

so that's (750^2 -z^2)-2*10^3/3

I'm not sure why you did that... but if you want to make it 'not wrong', it should be equal to zero.

... 750^2-2000/3= 56183

I know something is wrong with the answer.. it looks funny

Can you try again with the new equation you got? :)

cassi999 · Oct 18, 2011, 06:33 AM

okay so basically √(750^2-(2000/3)) = 749.5
Is that correct?
Thank you

Unknown008 · Oct 18, 2011, 10:30 AM

Yes yes it is! :)

Okay, can you try the next one? Post what you came up with, just like the previous.

cassi999 · Oct 22, 2011, 05:39 AM

hi I have done transpose could you check to see if it is correct
u=v-at where a=5 t=10 and u=25
2) s=1/2(v+u)t where s=1550 u=75 and t=6.1
3) h= v^/ 2g where h=2.5*10^6 and g=9.81
are the questions

answers
u = v - at
v = u + at
= 25 + 50 = 75

s = 1/2(v + u)t
2s/t = v + u
v = ((2s)/t) - u
= (3100/6.1) - 75

h = v^2/2g
v^2 = 2hg
v = sqrt(2hg)
= sqrt(2 * 2.5 * 10^6 *9.81)

I didn't understand this one though
4) x=¹(2yv-z) where x=2.5*10-^3 and z=1.2*10-^6

Unknown008 · Oct 22, 2011, 07:04 AM

Good! Very good even! :)

And... the last one didn't come out well... there's a weird symbol in there :(

cassi999 · Oct 23, 2011, 03:52 PM

Can u go through with the last question Thank you

Unknown008 · Oct 24, 2011, 12:08 AM

I would be glad to, but as I told you there is this "¹" symbol which I don't know what that means...

cassi999 · Oct 25, 2011, 01:20 PM

Hi
I can't see that weird a symbol do u mind showing ot to me please
Thanks

Unknown008 · Oct 26, 2011, 02:55 AM

Can you see that?

cassi999 · Oct 26, 2011, 01:33 PM

u^2+ v^2-u^2=2a(ut+1/2 at^2)+u^2
I can not use the cancel out sign.. n could not copy n paste it from word.. but when u^2 is added in both sides you basically cancels out u^2 from each side... the when you said square root it.. do you then make v2 as the subject? For e.g. v^2 = √-u^2=2a √(ut+1/2 at^2) is that right

Thank you

Unknown008 · Oct 27, 2011, 08:15 AM

Okay, now I'm confused at what you're asking for.

For this:

$v^2 - u^2 = 2a$ut+\frac12 at^2$$

You did well with adding u^2, though it would be better to put in on the right of the terms, like this:

$v^2 - u^2 +u^2= 2a$ut+\frac12 at^2$ +u^2$

It doesn't change anything, but it helps you to see what is there a little better, I think.

This now gives:

$v^2 = 2a$ut+\frac12 at^2$ +u^2$

Take the square root on both sides, right.

$\sqrt{v^2} = \sqrt{2a$ut+\frac12 at^2$ +u^2}$

$v = \sqrt{2a$ut+\frac12 at^2$ +u^2}$

cassi999 · Oct 27, 2011, 10:33 AM

hi

its x=√(2yv-z) where x=2.5 *10^-3 y=35-10^-3 and z=1.2-10^-6.

for this question v^-u^/2a= u t 1/2 a t^

i have added u^2 for both sides but i didn't know what to do to v^2 to make it the subject it was confusing i got 57.22 is it correct or have i done it wrong thanks

Unknown008 · Oct 27, 2011, 10:48 AM

I'm getting something else. Did you see my above post when it came to v?

$v = \sqrt{(2(0.75)((3.6)(10.5) + (0.5)(0.75)(10.5^2))+3.6^2}$

This does not get me 57.22...

Okay, I give up on trying to understand the symbol the other one. But I guess it's some square root?

$x = \sqrt{2vy - z}$

Start by squaring both sides.

cassi999 · Oct 27, 2011, 02:53 PM

Yes it's a square root sign I don't know why it keeps doing that.
I have squared rooted it and got 11.47

cassi999 · Oct 27, 2011, 04:49 PM

hi
X² = 2YV – Z
2YV = X² + Z
V = (X² + Z) / 2Y
is the answer 4.11