# I NEED URGENT some math help u.u 13 replies

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Silberio VIP Member

Bourée

392,819 XP

9th October 2007

37,218 Posts

#1 6 years ago

Hey guys... I'm sorry to bother ya'll with this, but I'm really between the wall and the sword, as we say in Chile.

I really need urgent help with something. I've got homework for tomorrow and I can't find out a few things on it.. It's linear equations, AFAIK, and it's high school-level math, so it's not really THAT hard and shouldn't require extensive algorythms... I'm just to stupid to get it in my head.

1st problem:

Carly's got a telefone subscription which costs \$59 monthly and \$0.69 per minute call-fee. For Sam, the fee is \$99 monthly and \$0.49 per minute.

a. What question can you answer by solving the equation 59 + 0.69t > 99 + 0.49t?

b. solve the equation and answer the question.

2nd, true or false:

If a > b, then -a > -b

The equation x^2 = a has always a positive and a negative root.

You can not get (or take) the cube root of a negative number

Now, I've got pretty much half of the test and this is the other half... I'd greatly apriciate thine help, guys! :(

SeinfeldisKindaOk

5.56 smoke Haji every day

55 XP

18th July 2008

8,397 Posts

#2 6 years ago

42

SuperSmeg

50 XP

22nd September 2005

10,694 Posts

#3 6 years ago

This many!

*holds up 11 fingers*

Silberio VIP Member

Bourée

392,819 XP

9th October 2007

37,218 Posts

#4 6 years ago

AH DAMnIT :bawl:

Totes

Misanthrope

92,890 XP

7th January 2011

8,664 Posts

#5 6 years ago

Chuck Norris.

217,010 XP

7th December 2003

20,003 Posts

#6 6 years ago

a) you can determine the time necessary for plan a to be more expensive than plan b. In other words, does Cathy blabber long enough to get her money's worth or should she get the same plan as that other dude?

59 + 0.69t > 99 + 0.49t 0.69t-0.49t>99-59 0.2t>40 t>200 -> if t is larger than 200 plan a is more expensive than plan b. If it is equal 200 both plans cost the same, if it is smaller than 200 plan b is more expensive than plan a

solving an equation for one unknown factor isn't so hard. Don't be confused by the ">,<" signs, just reat them same as the "=" sign and keep in mind that there are some special rules such as the "<" sign turning around if you multiply with -1

2a: false. If you multiple a>b with -1 the ">" sign turns around. Example: 3 is larger than 2, but minus 2 is larger than minus 3. 2b: true, I guess, as long as a is >0. This equation has two solutions, one positive and one negative. But usually you discard the negative solution as a root of a negative number is not defined. If a is 0 the equation has one solution (or maybe it is not defined, can't remember), if a is <0 the equation has no solution. 2c: yes and no. You can't take the root of a negative number unless you use complex numbers.

Silberio VIP Member

Bourée

392,819 XP

9th October 2007

37,218 Posts

#7 6 years ago

Thank you so very much, MrFancypants.

As I see, I was kind of right on 2a as I had guessed it... I just didn't know about the -1 that turns around the ">" sign.

And I still don't really get the first one, but I understand the equation... But thanks a lot!

Schofield VIP Member

om :A

319,554 XP

24th October 2007

30,539 Posts

#8 6 years ago

MrFancypants;5602100a) you can determine the time necessary for plan a to be more expensive than plan b. In other words, does Cathy blabber long enough to get her money's worth or should she get the same plan as that other dude?

59 + 0.69t > 99 + 0.49t 0.69t-0.49t>99-59 0.2t>40 t>200 -> if t is larger than 200 plan a is more expensive than plan b. If it is equal 200 both plans cost the same, if it is smaller than 200 plan b is more expensive than plan a

solving an equation for one unknown factor isn't so hard. Don't be confused by the ">,<" signs, just reat them same as the "=" sign and keep in mind that there are some special rules such as the "<" sign turning around if you multiply with -1

2a: false. If you multiple a>b with -1 the ">" sign turns around. Example: 3 is larger than 2, but minus 2 is larger than minus 3. 2b: true, I guess, as long as a is >0. This equation has two solutions, one positive and one negative. But usually you discard the negative solution as a root of a negative number is not defined. If a is 0 the equation has one solution (or maybe it is not defined, can't remember), if a is <0 the equation has no solution. 2c: yes and no. You can't take the root of a negative number unless you use complex numbers.

wat

Jeff is a mean boss

565,401 XP

28th July 2002

53,121 Posts

#9 6 years ago

I like nachos

Guest

I didn't make it!

0 XP

#10 6 years ago

Fancy beat me :bawl:

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