CPhill

And if we wanted an "exact" value...it would be... 6√2.....

√(2v-7)=v-3     .... square both sides to get rid of the radical

2v - 7  = v^2 - 6v + 9    rearrange

v^2 - 8v + 16   = 0        factor

(v - 4)^2  = 0      and it's obvious that 4 is our solution

Make sure you check the solutions to this type of problem......the squaring of both sides sometimes results in "extraneous" solutions that don't "work" in the original equation.....fortunately, this one is OK.....!!!

Ah.....a little tricky, aren't we???  I forgot about those "leftover" butts.....!!

{I would have failed this one....LOL!!}

Here you go......

I've recommended that one for our Daily Wrap.......a very gpod effort, indeed!!

-18 - [75 / -15] =

-18 - [-5] =

-13

sin^2(x)+cos^2(x)+tan^2(x)+cot^2(x)+sec^2(x)+csc^2(x)=7

Note that  ...sin^2x + cos^2x = 1    ...so we have

tan^2x + cot^2x + sec^2x + csc^2x = 6

And

1 + tan^2x = sec^2x   ....so we have

2tan^2x + cot^2x + csc^2x = 5

And 1 + cot^2x = csc^x     so we have

2tan^2x + 2cot^x  = 4    divide through by 2   

tan^2x + cot^2x = 2

So

sin^2x/ cos^2x + cos^2x/sin^2x = 2

[sin^4x + cos^4]  = 2sin^2xcos^2x    rearrange

sin^4x - 2sin^2xcos^x + cos^4x = 0   factor

(sin^2x -cos^2x)(sin^2x - cos^2x)  =0   setting one of the factors to 0, we have

sin^2x - cos^2x = 0   factor again

(sinx + cosx)(sinx-cosx)  = 0

For the first factor, we have sinx = -cosx.... and this occurs at 3pi/4 and at 7pi/4

For the second factor, we have  sinx = cos x  and this occurs at pi/4 and 5pi/4

So....our solutions are   pi/4, 3pi/4, 5pi/4 and 7pi/4

If the area of circle B = 64*pi cm^2, it's radius must be 8 cm.

Then, the radius of circle A must be 1/2 of that = 4cm

And it's area = pi*(4cm)^2  = 16pi cm^2

OK.....post your quesion....