All Answers 356000 Answers

Thank you. I couldn't figure it out on my own.

32/42 x 100% = ?

48 m / 12 m/g   +  48 m / 24 m/g = 4 + 2 = 6 g

Total miles 96 miles     96 m / 6g = 16 m/g

Slope is  rise / run   or   (y₁-y₂₎ / (x₁-x_{2)      Note that it does not matter which point you call  1 and which you call 2}

Can you solve this now? 

Thank you heureka!

See http://mathworld.wolfram.com/ParametricEquations.html

There it is explained.

  !

Yes I saw that too.

I am hoping that ALL the other answerers saw it as well!

Yes, I see your point. The AoPS cheating issue is a problem, and has been a problem for at least three years.  I’ve noticed Wonderman’s posts tagging AoPS’ homework questions, and I’ve read the hot debates among members and guests precipitated by this, along with the comments of one or more trolls directed to particular offenders. Some of these were quite funny!

This is very funny!

A particle is moving so that its position at time  is given by the parametric equations  \(\begin{align*} x &= 5\sin(-2t) \\ y &= 5\cos(2t). \end{align*}\)
What is the speed of the particle?

\(\begin{array}{|rcll|} \hline \begin{align*} x &= 5\sin(-2t) & y &= 5\cos(2t) \\\\ \dot{x} &= (-2)\cdot 5\cos(-2t) & \dot{y} &= 2\cdot 5\cdot (-\sin(2t)) \\ &= -10\cdot \cos(-2t) & &= -10\cdot \sin(2t) \\ &= -10\cdot \cos(2t) \\ \end{align*} \\ \hline \end{array} \)

speed

\(\begin{array}{|rcll|} \hline v(t) &=& \sqrt{\dot{x}^2+\dot{y}^2} \\ &=& \sqrt{\Big(-10\cdot \cos(2t)\Big)^2+\Big(-10\cdot \sin(2t)\Big)^2} \\ &=& \sqrt{100\cdot \cos^2(2t)+100\cdot \sin^2(2t) } \\ &=& 10\cdot \sqrt{\cos^2(2t)+\sin^2(2t) } \quad | \quad \cos^2(2t)+\sin^2(2t) = 1 \\ &=& 10\cdot 1 \\ \mathbf{v(t)} &=& \mathbf{10} \\ \hline \end{array} \)

Hi doorknob,

Tomsun was trying to help you, he/she obviously did not know what complex numbers are. 

That would be true of most people here.

Please do not be rude to people who take an interest in your questions.

You could have just said something like,

"Hi Tomsun,

4i is a complex number, you will probably learn about those in the future."

That would have sounded much nicer.

w and z are 2 different complex numbers.

|wz|=40

|w+z|=13

w = 3+4i

find z

\(\text{Let $z=x+yi$}\)

1.

\(\begin{array}{|rcll|} \hline \mathbf{|wz|} &=& \mathbf{40} \\ \hline wz &=& (3+4i)(x+yi) \\ &=& 3x+3yi+4xi-4y \quad | \quad i^2 = -1 \\ &=& 3x-4y+(4x+3y)i \\ |wz| &=& \sqrt{(3x-4y)^2+(4x+3y)^2} \\ &=& \sqrt{ 9x^2-24xy+16y^2+16x^2+24xy+9y^2 } \\ &=& \sqrt{ 25x^2 +25y^2} \\ |wz| &=& 5\sqrt{ x^2 +y^2} \quad | \quad |wz| = 40 \\ 40 &=& 5\sqrt{ x^2 +y^2} \quad | \quad : 5 \\ 8 &=& \sqrt{ x^2 +y^2} \\ \sqrt{ x^2 +y^2} &=& 8 \\ \mathbf{x^2 +y^2} &=& \mathbf{64} \qquad (1) \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline \mathbf{|w+z|} &=& \mathbf{13} \\ \hline w + z &=& (3+4i)+(x+yi) \\ &=& (3+x) + (4+y)i \\ |w+z| &=& \sqrt{(3+x)^2+(4+y)^2 } \\ &=& \sqrt{ 9+6x+x^2+16+8y+y^2 } \\ &=& \sqrt{ 25+6x+8y +x^2+y^2 } \quad | \quad x^2+y^2 = 64 \\ &=& \sqrt{ 25+6x+8y + 64 } \\ |w+z| &=& \sqrt{ 89+6x+8y } \quad | \quad |w+z| = 13 \\ 13 &=& \sqrt{ 89+6x+8y } \\ 13^2 &=& 89+6x+8y \quad | \quad -89 \\ 80 &=&6x+8y \quad | \quad : 2 \\ 40 &=& 3x+4y \\ 4y &=& 40-3x \\ \mathbf{y} &=& \mathbf{\dfrac{40-3x }{4}} \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x^2 +y^2} &=& \mathbf{64} \quad | \quad \mathbf{y=\dfrac{40-3x }{4}} \\\\ x^2+\left( \dfrac{40-3x }{4} \right)^2 &=& 64 \\ x^2+ \dfrac{(40-3x)^2 }{16} &=& 64 \quad | \quad \cdot 16 \\ 16x^2+ (40-3x)^2 &=& 1024 \\ 16x^2+40^2-240x+9x^2 &=& 1024 \\ 25x^2-240x +576 &=& 0 \\\\ x &=& \dfrac{240 \pm \sqrt{240^2-4\cdot 25\cdot 576} }{2\cdot 25 } \\ x &=& \dfrac{240 \pm \sqrt{57600 -57600} }{50} \\ x &=& \dfrac{240}{50} \\ \mathbf{x} &=& \mathbf{\dfrac{24 }{5} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{\dfrac{40-3x }{4}} \quad | \quad \mathbf{x=\dfrac{24}{5}} \\\\ y &=& \dfrac{40-3\cdot \dfrac{24}{5} }{4} \\ y &=& 10-3\cdot \dfrac{6}{5} \\ y &=& \dfrac{50-18}{5} \\ y &=& \dfrac{32}{5} \\ \mathbf{y} &=& \mathbf{\dfrac{32}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{z} &=& \mathbf{\dfrac{24}{5} + \dfrac{32}{5}i } \\ \hline \end{array}\)

Thanks Ginger, but AoPS protect their images for a reason.

I know a lot of kids cheat on here, or go against the instructions of their educators by gaining outside help. (not always the same thing in my view)

But I am not going to purposefully circumvent the protection that educators are trying to put in place. :)

As always though, I am impressed by your knowledge and am grateful for your attempt to help.   

AoPS protects their images with an obfuscation code that’s active when viewed directly.

This obfuscation is easy to remedy:

For Chrome broswers: Right click on the page and select inspect. Note the three lines of code.

One of the lines will look like this:  ody style="margin: 0px; background: #0e0e0e;" >

Right click on the element and select edit. Edit the hex code to “ffeeee” (no quotes), then press enter. Close the inspection dialogue box.

The page will now be easy to view. 

GA

                .

This response shows how much your learn from us.

AND this is on one of your better days.

What is the smallest whole number that has

a remainder of 1 when divided by 4,

a remainder of 1 when divided by 3,

and a remainder of 2 when divided by 5

So it becomes:

x = 1 mod(4)

x = 1 mod(3)

x = 2 mod(5)

\(\begin{array}{rcll} x &\equiv& {\color{red}1} \pmod {{\color{green}4}} \\ x &\equiv& {\color{red}1} \pmod {{\color{green}3}} \\ x &\equiv& {\color{red}2} \pmod {{\color{green}5}} \\ \text{Let } m &=& 4\cdot 3\cdot 5 = 60 \\ \end{array}\)

Because 4 and 3 and 5 are relatively prim  we can go on:

\(\begin{array}{|rclcl|} \hline x &=& {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot \left[\dfrac{1}{\color{green}3\cdot 5}\pmod{4} \right] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot \left[\dfrac{1}{\color{green}4\cdot 5}\pmod{3} \right] \\ &+& {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot \left[\dfrac{1}{\color{green}4\cdot 3}\pmod{5} \right] + {\color{green}4\cdot 3\cdot 5}\cdot k \quad | \quad k \in \mathbb{Z} \\ \hline && \left[\dfrac{1}{\color{green}3\cdot 5}\pmod{4} \right] \\ && \equiv \left[ { (\color{green}3\cdot 5) }^{\varphi({\color{green}4}) -1 } \pmod {{\color{green}4}} \right] \quad | \quad \varphi({\color{green}4}) = 2 \\ && \equiv \left[ { 15 }^{2 -1} \pmod {{\color{green}4}} \right] \\ && \equiv \left[ { 15 } \pmod {{\color{green}4}} \right] \\ && \equiv \left[ { 3 } \pmod {{\color{green}4}} \right] \\ \hline && \left[\dfrac{1}{\color{green}4\cdot 5}\pmod{3} \right] \\ && \equiv \left[ { (\color{green}4\cdot 5) }^{\varphi({\color{green}3}) -1 } \pmod {{\color{green}3}} \right] \quad | \quad \varphi({\color{green}3}) = 2 \\ && \equiv \left[ { 20 }^{2 -1} \pmod {{\color{green}3}} \right] \\ && \equiv \left[ { 20 } \pmod {{\color{green}3}} \right] \\ && \equiv \left[ { 2 } \pmod {{\color{green}3}} \right] \\ \hline && \left[\dfrac{1}{\color{green}4\cdot 3}\pmod{5} \right] \\ && \equiv \left[ { (\color{green}4\cdot 3) }^{\varphi({\color{green}5}) -1 } \pmod {{\color{green}5}} \right] \quad | \quad \varphi({\color{green}5}) = 4 \\ && \equiv \left[ { 12 }^{4 -1} \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 12^3 } \pmod {{\color{green}5}} \right] \quad 12 \pmod{5} \equiv 2 \pmod{5} \\ && \equiv \left[ { 2^3 } \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 8 } \pmod {{\color{green}5}} \right] \\ && \equiv \left[ { 3 } \pmod {{\color{green}5}} \right] \\ \hline x &=& {\color{red}1} \cdot {\color{green}3\cdot 5} \cdot [3] + {\color{red}1} \cdot {\color{green}4\cdot 5} \cdot [2] + {\color{red}2} \cdot {\color{green}4\cdot 3} \cdot [3] + 60\cdot k \quad | \quad k \in \mathbb{Z} \\ x &=& 45+ 40 + 72 + 60\cdot k \\ x &=& 157+ 60\cdot k \quad | \quad 157 \pmod {60} = 37 \\ x &=& 37+ 60\cdot k \qquad k \in Z \\ \mathbf{x_{min}} &=& \mathbf{ 37} \\ \hline \end{array}\)

YES YOU  DO!!      

The other method is the stars and bars method

There are 6 stars, they are seperated into 3 groups so we need 2 bars to seperate them

One group is before the bars, one is after and the other is inbetween

so that is 8 'things' and I need to find how many different places the bars can go.

That is

8C2=28

How many ways are there to put 6 balls in 3 boxes if the balls are not distinguishable but the boxes are?

Number in the boxes

6,0,0             3 ways

5,1,0              6 ways

4,2,0              6 ways

4,1,1              3 ways

3,2,1              6 ways

3,3,0              3 ways

2,2,2               1 ways

I get 28 ways

I just get a black screen :(

And he is still a very handsome turkey LOL.   

and you don't need to go on my every post to say the same thing... telling all that to people helping me to get my stuff done

Hey man... i'm not very good with math and i don't treat nobody as my servants to do everything for me, you got that twisted