Melody

Do they get any interest on the money that they save?

Maybe ths will help both of you :)

((-1*0.056)-√(0.056²-(4*1*0.056)))/2)

((-0.056)-√(0.003136-0.224))/2)

(0.003136-0.224) is less than 1 and there are no real solutions to the square root of a negative number.

so

This has no real solutions

It has complex solutions but they are imaginary numbers so you don't need to worry about those. :)

How do I center a small circle into a bigger one perfectly...

The equation of a circle is

\(\boxed{(x-h)^2+(y-k)^2=r^2}\)where (h,k) is the centre and r is the radius.

so if  you have 2 circle equations with the same centre but with different radii, they will be concentric circles.

eg

\((x+1)^2+(y-3)^2=16\qquad and\\ (x+1)^2+(y-3)^2=9\)

Please Lisa can you be careful to include ALL the information in your original question AND be careful to include all neccessasry brackets as well as any extra brackets that will help people to understand your question properly in the first place.

Otherwise people waste a lot of their time and next tme they may be less likely to offer you any help at all.

Thankyou everyone for your great answers :)

question:  I still don't get it.

3+1=4

and

3-1=2

y=2 and y=4 are on either side of the centre - the middle is y=3   

so (0,3) must be the 'centre' of your graph,

if the graph was   y = tanx then one centre  is (0,0)  

If you want the centre to be (0,3) it will have to be  y= (tanx) +3

does that help?

Ninja's question is

"Where the heck did you get the 3 from"    :D

y=tanx   is centred on (0,0)     (at least on of the periods is)

so if you wanted points rotationally symmetrical on either side  they would be    

(0-x,0-y) and (0+x,0+y)

so they could be

your points are   (-pi/2 , 2) and (pi/2 , 4)

The y values are 2 and 4 so the centre of them is 3  Hence the graph has be lifted up 3 .

That is where the 3 came from :/

log3(x+5)=1-log3(x+3)

Thanks guest, I just want to look too :)

\(log_3(x+5)=1-log_3(x+3)\\ log_3(x+5)+log_3(x+3)=1\\ log_3[(x+5)(x+3)]=1\\ log_3[(x+5)(x+3)]=log_33\\ (x+5)(x+3)=3\\ x^2+8x+15=3\\ x^2+8x+12=0\\ (x+6)(x+2)=0\\ x=-6 \qquad or \qquad x=-2\\ \text{But you cannot find the log of a negative number so -6 is no good}\\ x=-2\\ check\\ LHS=log_33=1\\ RHS=1-log_31=1\\good:)\)

x=-2 just as our first guest already determined. :)

Since it is tan maybe this is what is wanted.  It still doesn't make much sense though.

y=3+tan(x/2)

Oh dear, I did not see the tangent function part.