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^

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? :)

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

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.

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

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? :)

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

Yes yes it is! :)

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

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

Good! Very good even! :)

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

Can u go through with the last question Thank you

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

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

Can you see that?

http://p1cture.it/images/68bbf1f8c7df29e1b2e9.png

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

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}

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

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.

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

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

I don't know it keeps showing me these weird symbols its suppose to be

x^2=2yv-z
2yv=x^2+z
v=(x^2+z/2y
4.11

1. Yes, the answer I got is 11.47.

2. No, I'm getting something totally different here... though you worked it out well.

v = \frac{x^2 + z}{2y}

v = \frac{(2.5\times10^{-3})^2 + (1.2\times 10^{-6})}{2(3.5\times10^{-3})}

That would give me 0.00017 as the answer... Is that right

Hmm, I'm sorry, but this is not what I am getting, unfortunately.

v = \frac{6.25\times 10^{-6} + 1.2 \times 10^{-6}}{7.0\times 10^{-3}}

All right? A little easier like that?

v = \frac{7.45\times 10^{-6}}{7.0\times 10^{-3}}

okay 7.45/7.0 then you subtract the powers so it becomes 10^-3... which would give you 1.0642 *10^-3 =0.001064 is that right ?

That's what I got :)

hi
can you check this for me
4ac^2=c^2+3ab when a=4.5 and b=3.6. make c the subject

4ac^2=c^2+3ab when a=4.5 and b=3.6
4*4.5*c^2 = c^2 + 3*4.5*3.6
18c^2 = c^2 + 48.6
17c^2=48.6
c^2 = 48.6/17

c = +/- sqr(48.6/17) = +/-1.6908

Yea you can do it like that, but this defeats the purpose of making c the subject... :(

Start by moving c^2 on the left side, then factor c^2.

okay so
4ac^2 = c^2+3ab
4(4.5)c^2 = c^2 + 3(4.5)(3.6)
18c^2 = c^2 + 48.6
17c^2 = 48.6
c^2 = 48.6/17
c^2 = 2.86
c = 1.69

No no no. You don't put in the numbers until you have c as the subject.

4ac^2 = c^2 + 3ab

4ac^2 - c^2 = 3ab

Forget the values of a and b for the time being and make c the subject of formula.

c^2=4ac^2+3ab is this arrangement right ?

No, you have c^2 on both sides...

Take from the last line in my previous post. Factor out c^2 just like you would factor x here:

ax - 2x = x(a - 2)

hi
I tried it and this is what I got. :-)

4ac² - c² = 3ab

=> c²(4a - 1) = 3ab

=> c² = 3ab/(4a - 1)

c = √(3ab/(4a - 1))

4ac² - c² = 3ab

=> c^2(4a - 1) = 3ab

=> c^2 = 3ab/(4a - 1)

c = squr(3ab/(4a - 1))

sorry it suppose to be
4ac^2 - c^2 = 3ab
I don't know why it keeps showing these weird symbols

Type them manually instead of copy/pasting from Microsoft word or something, but if I understood it well, you did a good job! Replacing the values of a and b in the last equation you got, that is:

c = \sqrt{\frac{3a}{4a-1}}

You should get c as 1.69 :)