1,000,000 Question Thread :D

[kenhung123]

(/inty,5] closed circle?

[TrueTears]

Maybe you forgot the negative. But otherwise yes, it is:



So using the same principle you can work out the domain of the line on the RHS.

[kenhung123]

So it is [5,/inty)?

[TrueTears]

So it is [5,/inty)?

Correct


So now if you take the union of those 2 what do you get?



Which is the 'overall' domain of those 2, or more formally, the domain of the hybrid function.

[kenhung123]

Thx!

[kenhung123]

The length of a rectanle is 4cm more than the width. If the length were to be decreased by 5cm and the width decreased by 2cm, the perimeter would be 18cm. Calculate the dimensions of the rectangle.

l=w+4
l-5=w+4-2

l-5=w+2
2(l-5)+2(w+2)=18

l=19/2cm and w=5/2cm is this correct?

[/0]

Um...I don't get that answer.
My equations are



and

[kenhung123]

So you disregarded the previous statement?

[m@tty]

No after getting to the equation /0 has you sub in and solve.

[/0]

So you disregarded the previous statement?

Another way to think of it, let L' and W' be the new lengths after reduction.
L' = L - 5
W' = W - 2

The new perimeter is given by

2L'+2W' = 18

2(L-5)+2(W-2)=18

However, the equation L = W+4 still holds, since that was before the reduction in dimensions.

Using these two equations  you can solve for L and W.

[kyzoo]

kenhung I answered this question in your other thread. And I don't think you need simultaneous equations at all.

[kenhung123]

The length of a rectanle is 4cm more than the width. If the length were to be decreased by 5cm and the width decreased by 2cm, the perimeter would be 18cm. Calculate the dimensions of the rectangle.

l=w+4
l-5=w+4-2

l-5=w+2
2(l-5)+2(w+2)=18

l=19/2cm and w=5/2cm is this correct?

What is mathematically wrong in my equations? Why can't I carry on the information provided previously to construct my equation 2?

[Cataclysmic]

Look at your 1st equation. Say L = 10, then W = 6.
Your 2nd equation would say 5 = 8.
They've taken 5 from the length and 2 from the width.

Your EQ 1: L = W+4
Your EQ 2: l-5 = w+4 - 2
In equation 1, LHS = RHS, so if you minus 5 from the LHS, you need to -5 from the RHS for them to equal one another.

[kenhung123]

It says the length is 4cm more than width so L=W+4, the width requires 4cm more to be same as length?

[Ilovemathsmeth]

I think so based on only that info.