Melody

Yes the second answer is definitely the correct one. 

I am glad you gave it a go guest one, that is the best way to learn.  I hope you see this continuation.

Good work guest 2   

I shall also try to explain.

When the base-b number 11011_b is multiplied by b-1, then 1001_b is added, what is the result (written in base b)?

\(11011_b=b^4+b^3+0+b+1\\ (b-1)*(b^4+b^3+0+b+1)\\ =b^5+b^4+0+b^2+b\\ \quad\;\;-(b^4+b^3+0+b+1)\\ =b^5+0-b^3+b^2+0-1\\ Now\\ 1001_b=b^3+0+0+1\\~\\ \;\;\;(b^5+0-b^3+b^2+0-1)\\ \;\;\;+\qquad\;\;(b^3+0+0+1)\\ =b^5+0+0+b^2+0+0\\ =100100_b \)

Yes you are right :)  And so the answer shall be 2.25 

Add the two together and then substitute.

\(x+y=0.25y+0.25x+a+b\\ Sub\;\; in\;\; x=1 \;\;and \;\; y=2\\ 1+2=0.5+0.25+a+b\\ 3=0.75+a+b\\ 2.75=a+b\\ a+b=2.75\)

Hi Ginger,

LOL  This ant just made an understandable mistake.  No big deal.  :)

Find the absolute value of the difference of the solutions of \(x^2-5x+5=0\)

The solutions of any quadratic equation are

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\x=\frac{-b}{2a}\pm \frac{ \sqrt{b^2-4ac} }{2a} \)

So the ablosute value difference of the roots is

\(\text{abs value of diff of roots} \\ =\left|\left[ \frac{-b}{2a}+ \frac{ \sqrt{b^2-4ac} }{2a}\right] - \left[ \frac{-b}{2a}- \frac{ \sqrt{b^2-4ac} }{2a}\right] \right|\\ =\left|\left[ \frac{-b}{2a}+ \frac{ \sqrt{b^2-4ac} }{2a}\right] + \left[ \frac{b}{2a}+\frac{ \sqrt{b^2-4ac} }{2a}\right] \right|\\ =\left| \frac{ \sqrt{b^2-4ac} }{2a}+ \frac{ \sqrt{b^2-4ac} }{2a} \right|\\ =\left| \frac{ \sqrt{b^2-4ac} }{a} \right|\\ \)

for your question

\(x^2-5x+5=0\)

a=1    b=-5        c=5

\(\text{abs value of diff of roots }= \frac{ \sqrt{25-20}}{1}=\sqrt5\)

Is it?

I had to teach maths for nurses one time and they had all these super horrible questions involving rates.

I 'invented' this method all by myself. LOL

I have never seen anyone else use it in mathematics although it is very similar to methods regularly used in chemistry I think.

It makes difficult rates questions super easy!

\(2+3-4-1=0\)

\(\frac{2-1}{4-3}=1\\ 4-3-1+2=2\\ \frac{2+1}{4-3}=3\\ \frac{4}{2}+3-1=4\\ 4+3-1*2=5\\ \frac{4}{2}+1+3=6\\ (4+3)(2-1)=7\\ (3-1)^2+4=8\\ 1*2+3+4=9\\ 1+2+3+4=10\\ 2*3+4+1=11\\ 2^3+4*1=12\\ 2^3+4+1=13\\ 2*(3+4)*1=14\\ 3*(4+2-1)=15\\ 4*(3+2-1)=16\\ 3*(2+4)-1=17\\ 3*(2+4)*1=18\\ (2+3)*4-1=19\\ 1*(2+3)*4=20\)

Hi Vest4R,

\(3Tcos\theta - 1.5Tsin\theta = 0\\ T(3cos\theta - 1.5sin\theta )= 0\\ T=0\;\;or\;\; 3cos\theta - 1.5sin\theta=0\\ T=0\;\;or\;\; 3cos\theta = 1.5sin\theta\\ T=0\;\;or\;\; \frac{3cos\theta}{1.5cos\theta} = \frac{1.5sin\theta}{1.5cos\theta }\\ T=0\;\;or\;\; 2 = tan\theta\\ T=0\;\;or\;\; tan\theta=2\\ T=0\;\;or\;\; \theta=atan(2)\qquad\;\;NOTE: \theta\text{ must be in the first or third quad}\\ T=0\;\;or\;\; \theta=(63.43+180n)^\circ \qquad where \qquad n\in Z \)