this thread went way over my headâ€¦

i hate math. even stats

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hahahaha, nope definitely whent through all five years of high school. But in civil engineering in university now so math should technically be one of my things. Figured if I couldn't help her on that, I needed to change major asap :P

Glad you were able to understand though, wasn't sure if it was clear. Have an exam soon?

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Never heard of Khan though... What is it?

purplemath.com]]>

Nope i I volunteer to tutor math at a predominantly black high school. I've also helped other bhmers. I like helping niccas, with math Lol.]]>

and thanks for the Patrickjmt suggestion! Cool site!]]>

_ConcreteRose_ wrote:Yep Khan academy really helped me through calc 2. They have a whole bunch of material. Also look at PatrickJMT. Honestly whatever you'll work on is probably on you tube somewhere. |

Do you coach on Khan?

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Yep Khan academy really helped me through calc 2. They have a whole bunch of material. Also look at PatrickJMT. Honestly whatever you'll work on is probably on you tube somewhere.]]>

Erikaaa and Concrete, since you guys seem to be into math, have you ever heard of Khan Academy?]]>

Bored w/Out Me? wrote:WTG Erikaa, You graduate from High School @12??? |

lol

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WTG Erikaa, You graduate from High School @12???]]>

_ConcreteRose_ wrote:Good job! |

Thanks, girl!

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Good job!]]>

So when I finally get it I get a one star? Whoever you are, you're a funny girl! ]]>

Instead of just crossing stuff out I actually wrote the entire problem out step by step. I notice that whenever I try to skip a step or do it in my head, THAT'S when I mess up. By writing out each fraction and multiplying step by step I don't lose anything or forget what I was doing.]]>

Thank you to EVERYONE!!!!! that helped me. BHM has some smart and patient ladies! You're the best!

I do need to find some more problems like this though so I can get my technique down...

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Ahhhhhhh I got it after (several more attempts)!!!!!

Like others mentionned, the problem is you're multiplying the (x-y) by z twice since 5z*z(x-y) = 5 z^2 (x-y)

Basically, what you have to do is:

V = 5((x-y)/2) + w

Multiply both sides by z gives you:

V*z = Z * [5((x-y)/2) + w]

When you distribute the z to both the 5((x-y)/2) and the w, you get:

V*z = 5z((x-y)/2) + w*z. And not v*z = 5z*z((x-y)/2) + w*z (there's one too many z here as mentionned before)

Finally, when you eliminate the z in the first term on the right side of the equation, you get:

V*z = 5(x-y) + w*z

The z can't just be taken off since there is one term (the 5(x-y)) that doesn't have the z.

Both terms on the right side of the equation would have to have the z in them for you to be able to simplify your equation like you're trying to do,

But that isn't possible in this situation.

Makes sense? Sorry, which I could write it down and show you better than this...]]>

vz=5(x-y)+wz <------ correct answer

v=5(x-y)+w <---------your answer

What's the different here?

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_ConcreteRose_ wrote:
yes if you plan on multiplying something by it. |

Okay, okay, I'll try to let that sink in.

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_ConcreteRose_ wrote:edit. nvm |

No don't give up!

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liesnalibis wrote:
1(y) ----- =y 1 |

exaclty, now do the same thing with this fraction.

(z) (5(x-y))

--------

z

what are you left with?

Edited by _ConcreteRose_ - Dec 11 2013 at 11:18pm]]>

_ConcreteRose_ wrote:Exactly. Ok, Now solve this one (z)( y) --- z |

1(y)

----- =y

1

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liesnalibis wrote:
Ohhhhhhh so the 5(x-y) is considered a single expression? |

now i can go to sleep

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zolloh wrote:it's vz=5(x-y)+wz....not vz=5z(x-y)+wz it's vz=5(x-y)+wz....not vz=5z(x-y)+wz it's vz=5(x-y)+wz....not vz=5z(x-y)+wz |

Ohhhhhhh so the 5(x-y) is considered a single expression?

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liesnalibis wrote:
Ohhhhhhh so the 5(x-y) is considered a single expression? |

yes if you plan on multiplying something by it.

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edit. nvm

Edited by _ConcreteRose_ - Dec 11 2013 at 11:15pm]]>

_ConcreteRose_ wrote:Ok let me give you a problem 5 x 3 --- 5 How would you solve this? |

5 3 1 3 3

--- x --- = ---- x ---- = --- = 3

1 5 1 1 1

OR

5 3 15

--- x --- = ---- = 3

1 5 5

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Sang Froid wrote:Lie the z with the v and w don't cancel out. |

I don't understand...

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_ConcreteRose_ wrote:lmao. No. You multiplied the fraction by z twice. |

Can you explain where?

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it's vz=5(x-y)+wz....not vz=5z(x-y)+wz

it's vz=5(x-y)+wz....not vz=5z(x-y)+wz

it's vz=5(x-y)+wz....not vz=5z(x-y)+wz

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