math help.

[BmXStuD]

Simple highschool math but can u tell me pls the steps and answer of this problem.

p.s dont forgot to add the 1=5 ect steps pls i really really need this help ;0 ty everyone.

Problem

1. 4m+ (-3m) +2m

2. (6(small 3) +4z -5+(3z(small 2) -z +2


what i mean by small 3 and 2 is that they are expoents ;-0 not sure how to do all the small numbers on keyboard ù:
Also if a problem was like this "4x+Z +4x -Z" what does the Z mean is that a 1 or u just dont add or subtract anything?

 

[rival]

If you cant do that, you might as well drop the class and forget college

 

[dighn]

Originally posted by: BmXStuD
Simple highschool math but can u tell me pls the steps and answer of this problem.

p.s dont forgot to add the 1=5 ect steps pls i really really need this help ;0 ty everyone.

Problem

1. 4m+ (-3m) +2m

2. (6(small 3) +4z -5+(3z(small 2) -z +2


what i mean by small 3 and 2 is that they are expoents ;-0 not sure how to do all the small numbers on keyboard ù:
Also if a problem was like this "4x+Z +4x -Z" what does the Z mean is that a 1 or u just dont add or subtract anything?

for 1.
factor out the m and go from there
2. 6(small 3) should be written as 6^3 by convention.

yes Z = 1Z

 

[dighn]

Originally posted by: rival
If you cant do that, you might as well drop the class and forget college

he probalby has a really low grade level, give him a breka

 

[fornax]

1. SInce you have m and only m in all terms, you just add the coefficients of m:

4m + (-3m) + 2m = (4-3+2)m = 3m

Think of m as something, i.e. beans (That funny scene in Black Adder comes to mind, where he teaches Baldrick to count). So you have

four beans minus three beans plus two beans makes three beans

2. 6^3 + 4z - 5 + 3z^2 - z + 2

1st, gather together all like terms (i.e. terms that have the same power of z):

z squared: only one term, 3z^2
z to the 1st power: two terms, 4z and -z
z to the zeroth power (i.e. constants): 6^3, -5 and +2

3z^2 + (4z - z) + (216 - 5 +2) =

3z^2 + 3z + 213

That's the simplest it will get unless you want (or need) to factor it.

 

[Darien]

I'm not quite sure what you're solving for. do you just want the most simplest form? Are these equations equal to 0?


1. 4m + (-3m) + 2m

4m - 3m + 2m
3m

If you're having a hard time mixing symbols with numbers, you can factor out the m as follows

4m + (-3m) + 2m --> m[4 + (-3) + 2] --> m[4 - 3 + 2] = 3m



2. 6^3 + 4z - 5 + 3z^2 - z + 2

Rearrange the problem so it looks something like the form (ax^2) + (bx) + (c)
(3z^2) + (4z - z) + (2 + 6^3 - 5)

6^3 is 216, so

(3z^2) + (4z - z) + (2 + 216 - 5)
(3z^2) + (3z) + (213)

Notice that in each part you can factor out a 3
3[(z^2) + (z) + (71)]

3(z^2 + z + 71)



EDIT: argh, beaten by 3 minutes

 

[BmXStuD]

Originally posted by: fornax
1. SInce you have m and only m in all terms, you just add the coefficients of m:

4m + (-3m) + 2m = (4-3+2)m = 3m

Think of m as something, i.e. beans (That funny scene in Black Adder comes to mind, where he teaches Baldrick to count). So you have

four beans minus three beans plus two beans makes three beans

2. 6^3 + 4z - 5 + 3z^2 - z + 2

1st, gather together all like terms (i.e. terms that have the same power of z):

z squared: only one term, 3z^2
z to the 1st power: two terms, 4z and -z
z to the zeroth power (i.e. constants): 6^3, -5 and +2

3z^2 + (4z - z) + (216 - 5 +2) =

3z^2 + 3z + 213

That's the simplest it will get unless you want (or need) to factor it.

ty, and btw the asshole @ the top im not the best at math so stfu. i have to try this out @ skool 2marrow.

 

[BmXStuD]

EDIT: argh, beaten by 3 minutes

Its cool all the help i can get is cool to me. ;0

 

[johnjohn320]

Originally posted by: rival
If you cant do that, you might as well drop the class and forget college

Well as long as there are encouraging, helpful people like yourself, I'm sure he'll be fine.