All Answers 358089 Answers

Thanks, MPS    !!!!!

BTW.....welcome.....!!!

The red figure  will just have  the surface area  of a sphere  of  radius  r

So....its surface area  is     4pir^2

The height of the cylinder  = 2r

So......its surface area  is   [ 2pi (r^2)  +  2pi*r (2r) ]   =   [ 2pi(r^2)  + 4pir^2 ]

So   the cylinder has the greater surface area

it's around or is 1.06498652656

er, did you want the answer? \(9z(z^2-3z+3)\) is my answer.

-MPS101

Hi, when you multiply both sides by 10, you get \(5(r-45)=2(3-2r)\) because \(10 \div 2 = 5 \text{ and } 10 \div 5 = 2\) .

When you expand the equation you should get \(5r-225=6-4r\).

Moving the variables to one side and constants to another you get \(9r=231\).

Dividing both sides by 9 gets us \(r= 231\div 9\) which equals to \(25 \frac{2}{3}\).

-MPS101

(a) We want to distribute 50 votes among 4 categories.  The number of ways of doing this is \(\binom{49}{3} = 18424\)

(b) We want to distribute 50 votes among 5 categories, because we can think of abstaining as a category.  The number of ways of doing this is \(\binom{49}{4} = 211876\)

This is wrong

3z+4i\bar{z}=1-8i

\(3z+4i\bar{z}=1-8i\\ LHS=3(a+bi)+4i(a-bi)\\ LHS=3a+3bi+4ai+4b\\ LHS=3a+4b+(3b+4a)i\\ so\\ solve\;\;simultaneously\\ 3a+4b=1 \qquad and \qquad 4a+3b=-8\\ 12a+16b=4 \quad and \qquad 12a+9b=-24\\ subtract\\ 7b=28\\ b=7\\ so\\ 3a+28=1\\ 3a=-27\\ a=-9\\ so\\ a=-9\qquad b=7\\ z=-9+7i\)

You need to check it though.

Edit:

The final answer is incorrect because of a careless error near the end.   28/7 is 4

This error is of no significance. I am here to teach not not provide cheat sheets.

There is nothing wrong with my response. 

Especially given that I wrote it needed to be checked. But then all answers should be checked and fully understood.

You are welcome.  :)

Thank you, that was really useful!! 

Consider the trinomal 

\(2x^2-7x-4=0\\ \text{this factorises to }\;\;(2x+1)(x-4)=0\\ \text{so the roots }\alpha\;\;and\;\;\beta \;\; are \;\;\frac{-1}{2}\;\;and \;\;+4\\ \alpha+\beta=\frac{-1}{2}+4=3\frac{1}{2}=\frac{7}{2}\\ \alpha*\beta=\frac{-1}{2}*4=-2=-\frac{4}{2}\\ \\~\\ \)

Compare thes numbers to the coefficients in the original equation.  

What you can see is not a coincidence.

And it can be extanded to polynomials of ANY degree.

So in your polynomial

\(Ax^4+Bx^3+Cx^2+Dx^1+E=0 \)

If there are 4 real roots and they are

  \(\alpha,\beta,\gamma, \delta\;\;\;then\\ \alpha+\beta+\gamma+ \delta=-\frac{B}{A}\\ \text{The sum of the product of root pairs will be}\quad +\frac{C}{A}\\ \text{The sum of the product of root tripples will be}\quad -\frac{D}{A}\\ and\;\;lastly\\ \alpha\beta\gamma \delta=+\frac{E}{A}\\ \)

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I am not sure how it works if the roots are not real.  I can't remember thinking about that before.

Here is a video that covers some of this.

The figure to the right shows an irregular hexagon with six circles of radius 1, where the hexagon's vertices are the circles' centers. Find the sum of the areas of the black regions.

The circle above has center O and area 9*pi.  Triangle ABC is equilateral.  Find its area.

Hello Melody,

how did you know that a+b+y+o=7?? 

This polynomial is degree 4 NOT 5

Let the roots be

    \(\alpha,\beta,\gamma,\delta\\ then\\ \alpha+\beta+\gamma+\delta=-\frac{-7}{1}=7\\ \text{The sum of the product of pairs of roots is}\;+\frac{4}{1}=4\\ \text{The sum of the product of triples of roots is}\;-\frac{7}{1}=-7\\ \alpha\beta\gamma\delta=+\frac{-2}{1}=2\\ \)

\(\frac{1}{x_{1}} + \frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}\\ \)

Now make this one entire fraction and use the facts above to get  a value.

Base 3.  211.  In base 10 this is 2*3²+1*3+1 = 22

Base 4.  112.  In base 10 this is 1*4²+1*4+2 = 22

If x - a is a factor of x^3 - 3*a*x^2 + 2*a^2*x + b, then find the value of b.

Hello Guest!

\(\frac{x ^ 3 - 3 a x ^ 2 + 2 a ^ 2 x + b}{ x-a}\\ =\frac{x ^ 3 - 3 a x ^ 2 + 2 a ^ 2 x }{x-a} +\frac{b}{x-a}\\ -\frac{x ^ 3 - 3 a x ^ 2 + 2 a ^ 2 x }{x-a}=\frac{b}{x-a}\\ \color{blue}b=-x^3+3ax^2-2a^2x \)

\(\frac{x ^ 3 - 3 a x ^ 2 + 2 a ^ 2 x }{x-a}=-2 a x + x^2\)

\((x-a)\ is\ a\ Divisor\ of\ (x ^ 3 - 3 a x ^ 2 + 2 a ^ 2 x).\\ (x-a)\ is\ not\ a\ Divisor\ of\ (x ^ 3 - 3 a x ^ 2 + 2 a ^ 2 x\color{blue}+b)\ (b\neq 0).\\ (x-a)\ is\ a\ Divisor\ of\ (x ^ 3 - 3 a x ^ 2 + 2 a ^ 2 x\color{blue}+0).\\ so \)

\( b=0\)

  !

You should know, Mr. BB, that using “brovos” for a woman is often considered mockery. The correct form is “brave” and “brava” is the singular form.

Thanks for the help! I figured out the answer! :)

1/x_1 + 1/x_2 + 1/x_3 + 1/x_4 = -7.

z = 4 + 8i.

(a,b) = (4,-1).