Rom

what is (x+(3+2i)) times (x-(3+2i)) ?

note that

\((a+b)(a-b)=a^2 - b^2\) 

and we get

\( (x+(3+2i)) (x-(3+2i))= x^2 -(3+2i)^2\)

\((3+2i)^2 = (9-4)+12i=5+12i\)

\( (x+(3+2i)) (x-(3+2i)) = x^2-5 -12i\)

What is 17 3/4 minus 8 7/12

First I'd note that

\(\dfrac 3 4 = \dfrac 9 {12} > \dfrac 7{12}\) 

so

\(17 \dfrac 3 4 - 8\dfrac 7{12} = (17-8) + \left(\dfrac 3 4 - \dfrac 7{12}\right)= \\ \\ 9 + \left(\dfrac 9{12}-\dfrac 7{12}\right) = 9+\dfrac 2{12} = 9\dfrac 1 6\)

might it be that the horse is named Friday?

An open square-based container is made by cutting 4cm square pieces out of a piece of tinplate. If the volume of the container is 120cm^3, find the size of the original piece of tinplate

The dimensions of the box will be, w x w x h.

 Each side will be w x h

\(\text{side area } = w h = 4 \\ \\ h = \dfrac 4 w\)

The volume is given by

\(V = w^2 h = 120 \\ \\ w^2\left(\dfrac 4 w\right) = 120 \\ \\ 4w = 120 \\ \\ w = 30 \\ h = \dfrac 4 {30} = \dfrac 2 {15}\)

So the size of the original piece of tin plate that is the bottom is w x w, or 30x30 cm²

Suppose that f is a function such that 3f(x)-5xf(1/x)=x-7 for all non-zero real numbers x. Find f(2010)

\(3f(2010) - 5(2010)f\left(\dfrac 1 {2010}\right) = (2010)-7\)

\(3f(2010)-10050f\left(\dfrac 1 {2010}\right) = 2003\)

we can also write

\(3f\left(\dfrac 1 {2010}\right) - \dfrac {5}{2010}f(2010)=\dfrac 1{2010}-7=-\dfrac{14069}{2010}\)

\(let f(2010)=x, f\left(\dfrac 1 {2010}\right)=y\)

\(3x-10050y=2003\\ -\dfrac{5}{2010}x + 3y = -\dfrac{10469}{2010}\)

Solving this we get

\(x=2896\\\\ y = \dfrac{1337}{2010}\)

and thus

\(f(2010) = 2896\)

61 is prime so the only numbers that divide it are 1 and 61

How do I solve ln(x-9)^4=8?

\(\ln(x-9)^4=8\\ \ln(x-9)=\pm 8^{\frac 1 4} = \pm 2^{\frac 3 4} \\ x-9 = e^{\pm 2^{\frac 3 4}} \\ x = 9+e^{\pm 2^{\frac 3 4}}\) 

X2-4x+4=12

\(x^2-4x+4=12 \\ (x-2)^2 = 12\\ (x-2)=\pm \sqrt{12}\\ x=2\pm 2\sqrt{3}\) 

5 is in the 10,000 place.

 

Are all equilateral octagons equiangular? 

 

No.  Imagine the vertices of a regular octagon are hinged joints.  Now deform it so it's no longer equiangular.

Are all equiangular octagons equilateral?

 

No.  Make a regular octagon.  Now equally stretch any pair of opposite sides.