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Binomial Distribution Table:

Orange-Orange:  0.55 x 0.55  =  0.3025

Orange-Green:  0.55 x 0.45  =  0.2475

Green-Orange:  0.45 x 0.55  =  0.2475

Green-Green:  0.45 x 0.45  =  0.2025

g(x) = x² + 4  is  f(x) = x²  shifted up 4

g(x) = 5x²  is  f(x) = x²  with a vertical stretch

\(Find\ \overline{AE}.\)

Hello Guest!

\(\overline{BC}=\sqrt{2^2+3^2-2\cdot 3\cdot cos\ 60°}\\ \overline{BC}=\sqrt{10}\)

\(\frac{sin\ BCD}{2}=\frac{sin\ 60°}{\sqrt{10}}\\ sin\ BCD=\frac{2\cdot sin\ 60°}{\sqrt{10}}\\ BCD=33.211°\)

\(ACE=180°-90°-33.211°\\ ACE=56.789°\)

\(EAC=180°-60°-56.789°\\ EAC=63.211°\)

\(\frac{\overline{AE}}{sin\ 56.789°}=\frac{2}{sin\ 63.211°}\\ \overline{AE}=\frac{2\cdot sin\ 56.789°}{sin\ 63.211°}\)

\(\overline{AE}=1.8745\)

  !

Left 4 .

Ohhhh That (1+k) cancels!!

Thank you, Melody!

(Also that person was me... but I forgot I wasnt logged in)

You have been shown how to do these two or three times now...... try it yourself !

   I suspect your next post will be the same type of question with - 5x^2 + 10x - 5   .....   I am corrrect, yah?

         Do your own homework !

https://web2.0calc.com/questions/help_76862

The discriminant of the quadratic equation is:  b² - 4·a·c where a, b, and c

are the coefficients of the x²-term, the x-term, and the constant.

[provided that the equation is in this form:  ax² + bx + c  =  0, which it is]

In this case:  a = 2     b = 8    and     c = 8

Place these values into the discriminant and solve.

If the result is any negative number, the quadratic has two imaginary solutions.

If the result is zero, the quadratic has one real solution.

If the result is andy positive number, the quadratic has two real solutions.

Same problem here. I've been trying to register for weeks. I've tried everything from using a different network to clearing my cookies but nothing works.

https://web2.0calc.com/questions/reporting-problems

I already did that still did nothing 

ye its science... haven't heard of it?

You have   \(\displaystyle \textbf{v}=\left[\begin{array}{c} 2a+3b\\ -a+b \end{array}\right]\) and require that  \(\displaystyle P\, \textbf{v}=\left[\begin{array}{c} 2a \\ -a \end{array} \right]\, .\)

So, let  \(\displaystyle P=\left[\begin{array}{cc} r & s \\ t & u \end{array}\right]\) say, multiply out the rhs and equate components across the equation to find r, s, t and u.

You should find that \(\displaystyle P = \left[ \begin{array}{cc} 2/5 & -6/5 \\ -1/5 & 3/5 \end{array} \right]\, .\)

discriminant of the quadratic formula

b2 - 4ac 

8^2 - 4 ( 6)(3)

64 - 72

 - 8                  <===== negative disriminant means  two imaginary solutions

Thanks EP:

A note to the guest asker:

If you are having problems like this and you are registering an error, please indicate what part of the world that you live in.

or maybe

Gee you put so much effort into presenting that question well !

How could I possibly resist asnwering. 

2√5x+32=7

\(\qquad2\sqrt5 \;x+32=7\\ 2\sqrt5 \;x+32-32=7-32\\ \phantom{ wwwww }2\sqrt5 \;x=-25\\ \phantom{ wwwwwwww } x=\frac{-25}{2\sqrt5}\\ \phantom{ wwwwwwww } x=\frac{-25}{2\sqrt5}\times \frac{\sqrt5}{\sqrt5}\\ \phantom{ wwwwwwww } x=\frac{-25\sqrt5}{10}\\ \phantom{ wwwwwwww } x=\frac{-5\sqrt5}{2}\)

That is an interesting question.

I'll take a look

\(\frac{a}{b}=\frac{ak}{bk}=\frac{c}{d}\\ so\\ \frac{a+c}{b+d}=\frac{a+ak}{b+kb}=\frac{a(1+k)}{b(1+k)}=\frac{a}{b} \)

Are you happy with that?  Of course this only works if k is not equal to -1.