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i got 4/5 i am confused

\(z\overline{w} = 6 + 8i\\ \overline{z\overline{w}} = \overline{6+8i}\\ \overline{z}w = 6 - 8i \)

And then, we know that for any complex number z and w, \(z\overline{z} = |z|^2\) and \(|z||w| = |zw|\).

\(\quad|zw|^2 \\= |z|^2 |w|^2 \\= z\overline{z}w\overline{w} \\= (z\overline{w})(w\overline z) \\= (6 + 8i)(6 - 8i) \\= 10\)

\(\quad|z + w|^2 \\= (z + w)\left(\overline{z + w}\right) \\= (z + w)\left(\overline z + \overline w\right)\\ = z\overline z + w\overline w + z\overline w + w\overline z \\= |z|^2 + |w|^2 + z\overline w + \overline zw \\= 5^2 + 2^2 + (6 + 8i) + (6- 8i) \\= 41\)

I will leave the remaining ones to you.

\(\csc x \tan x = \dfrac1{\sin x} \cdot \dfrac{\sin x}{\cos x} = \dfrac1{\cos x} \)

\(\dfrac1{\cos x} = \dfrac{2\sqrt 3}3\\ \dfrac1{\cos x} = \dfrac{2}{\sqrt 3}\\ \cos x = \dfrac{\sqrt 3}{2}\\ x = \dfrac{\pi}{6} \text{ or } x = \dfrac{11\pi}{6}\)

If the line is tangent to the circle, then the line and the circle only touch at one point. That means if I solve for x( or y), there should only be one solution.

\(\begin{cases}x = 2y + 5\\x^2 + y^2 = 5\end{cases}\)

Substituting the first equation into the second, 

\((2y + 5)^2 + y^2 = 5\\ 4y^2 + 20y + 25 + y^2 = 5\\ 5y^2 + 20y + 20 = 0\\ y^2 + 4y + 4 = 0\\ (y + 2)^2 = 0\\ \boxed{y = -2}\)

As you see, it only has one solution. Therefore the line is indeed tangent to the circle.

Definitions: 

\(\text{Arithmetic mean} = \dfrac{a + b}2\\ \text{Geometric mean} = \sqrt{ab}\\ \text{Harmonic mean} = \dfrac2{\dfrac1a + \dfrac1b}\)

We get, by using the definitions,

\(\begin{cases}a + b = 20\\\dfrac1a + \dfrac1b = \dfrac1{180}\end{cases}\)

From equation (2), 

\(\dfrac{a + b}{ab} = \dfrac1{180}\\ \dfrac{20}{ab} = \dfrac1{180}\\ ab = 3600\\ \text{Geometric mean} = \sqrt{ab} = \sqrt{3600} = \boxed{60}\)

Let d be the common difference.

\(\quad\dfrac{D^2 - A^2}{C^2 - B^2}\\ = \dfrac{(A + 3d)^2 - A^2}{(A + 2d)^2 - (A + d)^2}\\ = \dfrac{((A + 3d) - A)((A + 3d) + A)}{((A + 2d) - (A + d))((A + 2d) + (A + d))}\\ = \dfrac{3\color{red}d (2A + 3d)}{\color{red}d(2A+ 3d)}\\ = 3\)

\(a^2 - b^2 = (a - b)(a + b)\).

By Law of indices, \(9x^2 = 3^2 x^2 = (3x)^2\), and \(64 = 8^2\).

We use the formula and substitute a = 3x, b = 8.

\(9x^2 - 64 = (3x)^2 - 8^2 = (3x - 8)(3x +8)\)

Because AB and BC are horizontal and vertical lines,

AB = 9 - (-6) = 15 units

BC = 6 - (-2) = 8 units

By Pythagorean theorem,

AC = \(\sqrt{8^2 + 15^2}\) = 17 units

The perimeter is 8 + 15 + 17 = 40 units 

\(z^2 (iz + 1) + i(iz + 1) = 0\\ (z^2 + i)(iz + 1) = 0\\ z^2 = -i \text{ or } z = \dfrac{-1}{i} = i\\ \)

For z² = -i, let \(z = re^{i\theta}\).

\(r^2 e^{2i\theta} = 1\cdot e^{3i\pi/2}\\ r = 1\\ \theta = \dfrac{3\pi}4 \text{ or } \theta = \dfrac{7\pi}4\)

Therefore the other two solutions are \(z = e^{3i\pi/4}\) and \(z = e^{7i\pi/4}\), which is \(-\dfrac{\sqrt 2}2 + \dfrac{\sqrt 2}{2} i\) and \(\dfrac{\sqrt 2}2 - \dfrac{\sqrt 2}{2} i\).

Graph: https://www.desmos.com/calculator/bpbp0kmds9

If A, B, C, D  are consecutive terms in an arithmetic progression,
what is the value of \(\dfrac{ D^2 - A^2 } { C^2 - B^2}\)
Assume \(B^2\) is not equal to \(C^2\).

AP:

\(\begin{array}{|rcll|} \hline A &=& a \\ B &=& a+d \\ C &=& a+2d \\ D &=& a+3d \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{ D^2 - A^2 } { C^2 - B^2}} &=& \dfrac{(D-A)(D+A)} { (C-B)(C+B)} \\\\ &=& \dfrac{(a+3d-a)(a+3d+a)} { \Big(a+2d-(a+d)\Big)(a+2d+a+d)} \\\\ &=& \dfrac{3d(2a+3d)} { (a+2d-a-d)(2a+3d)} \\\\ &=& \dfrac{3d(2a+3d)} { d(2a+3d)} \\\\ &=& \mathbf{3} \\ \hline \end{array}\)

\(f(x) = 1 - \dfrac 2{x + 1}\\ f(f(x)) = 1 - \dfrac2{1 - \dfrac2{x + 1} + 1} = 1 - \dfrac1{\dfrac x{x + 1}} = -\dfrac1x\\ f(f(f(x))) = -\dfrac1{\dfrac{x - 1}{x + 1}} = \dfrac{1 + x}{1 - x}\)

Now substitute x = 5.

Thank you, MaxWong

hi guest! i'm not really sure if you meant  \(5^{5x}+1=25^{2x}\) or \(5^{5x+1}=25^{2x}\), so I'll solve them both for you! :)

if you meant \(5^{5x}+1=25^{2x}\), then you can first start off by simplifying the equation.

\(5^{5x}+1=25^{2x} \)

\(5^{5x}+1=(5^2)^{2x}\)

\(5^{5x}+1=5^{4x}\)

this has equation has \(\boxed{\textbf{no solution}}\) 

if you meant \(5^{5x+1}=25^{2x}\), then you can start off by simplifying again.

\(5^{5x+1}=25^{2x}\)

\(5^{5x+1}=(5^2)^{2x}\)

\(5^{5x+1}=5^{4x}\)

since the bases are the same, we can solve for the exponents by making them equal.

\(5x+1=4x\)

\(x+1=0\)

\(\boxed{x=-1}\)

i hope this helped you! :)

First one    remember    rate x time = distance       or     distance/rate = time

45 km / 54 km/hr = .8333333 = 5/6 hr

he drove a total of    45 km  + 42 km = 87 km     in   5/6 hr + 35/60 hr

87 km /(5/6+ 35/60)hr  = 61.4  km/hr

Put x = 5   into the equation and compute    5-1  / 5+1   = 4/6     this is  f(5)

   now put   4/6 in the equation and compute    this will be  f f(5)

     finally take the last answer you derived and put it in for x in the equation   this will be   f f f(5)

Given that point O is the center of the circle and OA=5, OC=8, and angle ACO=30∘, find cos(angle BOA). 

Use the Law of cosines to calculate side AC and the angle OAB.

∠OAB = ∠OBA

∠BOA = 180º - ( ∠OAB + ∠OBA )  

Complete the square.

\(4x^2+5x+2\\ =4(x^2+\frac{5}{4}+\frac{1}{2})\\ =\boxed{4(x+\frac{5}{8})^2+\frac{7}{16}}\)

I'd start by multiplying both sides by (z+7)^2 and then solve it as a quadratic.

Clean up your question. Put your LaTex in a LaTex box  (found on the ribbon)

Here is a Venn Diagram of this situation:

2  4   8       common ratio, r = 2

3   8    13   common difference, d= 5

(3x - 8)(3x+8)

The side length of the triangle is 18/11.

(a) There are 66 ways to place the balls.

(b) There are 25 ways to place the balls.