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The following should help:

Expected value = $\frac{2}{6}\times 6+\frac{4}{6}\times 1-3=-\frac{1}{3}$

What is the value of $(1 + i)^5 - (1 - i)^5$?

$\begin{array}{|rcll|} \hline (1+i)^5-(1-i)^5 &=& \left((1+i)^2\right)^2(1+i)-\left((1-i)^2\right)^2(i-1) \\ &=& \left(1+2i+i^2\right)^2(1+i)-\left(1-2i+i^2\right)^2(1-i) \\ &=& \left(1+2i-1\right)^2(1+i)-\left(1-2i-1\right)^2(1-i) \\ &=& \left(2i\right)^2(1+i)-\left(-2i\right)^2(1-i) \\ &=& 4i^2(1+i)-4i^2(1-i) \quad | \quad i^2 =-1 \\ &=& -4(1+i)+4(1-i) \\ &=& -4-4i+4-4i \\ \mathbf{(1+i)^5-(1-i)^5} &=& \mathbf{-8i} \\ \hline \end{array}$

Such triangles are NOT possible!!!

For the given perimeter, the equilateral triangle has the largest area.

P = 3x (units)               A = 3x (square units)

x = 4√3 

Two triangles are similar. The smaller triangle has a perimeter of 2 units and an area of 2 square units. The perimeter of the larger triangle is 8 units. What is the area of the larger triangle?

Hello Guest!

$A_{sm}^2:A_{lg}^2=P_{sm}:P_{lg}$

$2^2:A_{lg}^2=2:8\\ 2A_{lg}^2=2^2\cdot 8\\ A_{lg}^2=\dfrac{2^2\cdot 8}{2}\\ A_{lg}= \sqrt{16}$

$A_{lg}=4$

The area of the larger triangle is 4 square units.

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You too! 

John bought some pens and pencils. After he sold 1/3 of the pens and 3/5 of the pencils, he had twice as many pens as pencils left. Given John bought 50 more pens than pencils, how many pens and pencils did John have altogether?

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John bought 50+x pens and x pencils.

(50 + x - 1/3(50 + x)) / (x + 3/5x) = 2

x = 250

50+x + x = 550

It's not 4b!!!

(2B + 2) + B + (2B - 7) = 180            B = 37

∠A = 76º

∠B = 37º

∠C = 67º

John can use $\frac{1}{7}$of his money to buy 2 rulers and 4 pens.

John can also $1 - \frac{2}{7} = \frac{5}{7}$of his money to buy 6 rulers, 7 books, and 12 pens.

We could also see that he can buy 6 rulers and 12 pens using $\frac{1}{7} \times 3 = \frac{3}{7}$of his money.

7 books cost $\frac{5}{7} - \frac{3}{7} = \frac{2}{7}$of his money.

$7 \times 3 = \boxed{21}$

4 rulers + 8 pens = 2/7

6 rulers + 12 pens = 3/7

7 books = 2/7

21 books = 6/7 

C : J = 6 : 11

(6x + 65) / (11x + 65) = 5 / 7

x = 10

Total marbles = 6x + 65 + 11x + 65 = 300

a = 4 b

c = 2b-7

b                               summed = 180

4b       +      2b-7       + b   = 180      solve for b      then a = 4b      and   c = 2b-7

Let

f(x) = 8x + 3 if x <= 3

f(x) = 16 - 2x if x > 4

Find the sum of all values of x such that f(x)=0

Hello Guest!

$f(x) = 8x + 3=0\\ \color{blue}x_1=-\frac{3}{8}\\ f(x) = 16 - 2x=0\\ \color{blue}x_2=8$

${\color{blue}The\ sum\ of\ all\ values}\ of\ x\ such\ that\ f(x)\ =\ 0\ is\\ x_1+x_2=-\frac{3}{8}+8\color{blue}=7\frac{5}{8}$

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If a and b are the roots of x^2 + x - 3, then find 4b^2 - a^3.

Hello Guest!

$x^2 + x - 3=0\\ x=-0.5\pm \sqrt{0.25+3}\\ x=-0.5\pm 0.5\sqrt{13}$

$a=-0.5+0,5\sqrt{13}\\ b=-0.5-0.5\sqrt{13}$

$\large4b^2 - a^3=19$

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There are 58 numbers.

First ...expand the given equation to    a x^2   + b x  + c      form

   then the maximum will occur at   x = - b/ 2a

       use this value of 'x' in the equation to calculate the max value of the function.......

d = dimes           value = 10d

q = quarters = 36-d      value = 25 (36-d)         summed , these two = 420

10 d         +       25(36-d)         = 420     solve for d = the number of dimes ...then quarters = 36 - d

I'm pretty sure this is $f(2007) = \frac{\log|2007| - \log\left|\frac{2006}{2007}\right| + \log\left|\frac{-1}{2006}\right|}{2} = \boxed{\log\left(2007/2006\right)}.$

Heyy! Happy birthday cal! Sorry i joined the party late! 

1 cent plus 5 cents + 10 cents = 16 cents     26 cents left and 9 coins left      On more dime    two more nickles and 6 more pennies

pennies   7

nickels     3

dimes       2   

x^2 > 1/2

x >  sqrt (1/2)     x < sqrt 1/2

sqrt .5 = .707  

from  .707 to 1     and from  -1 to -.707     is a total of .58579   out of a span of  2    = .293          29.3 %

https://web2.0calc.com/questions/polynomials_130#r1

https://web2.0calc.com/questions/help-proportion_1#r1

We get the maximum when w = x = y = z.  This give a maximum of 256/625.