# Algebra

posted by .

(8n6 - 6n - 5n4) + (2n4 + 2n6 + 3n)
I think I have this answer but am not sure if its

10n^6 - 3n^4 - 3n OR 10n - 3n^6 - 3n^4

I am pretty confident on the first answer. Any help is appreciated! Thanks!

• Algebra -

10n^6 -3n^4 + 3n

• Algebra -

Wouldn't the + 3n at the end be - 3n?

## Similar Questions

1. ### Math

what are the answers to the following questions?
2. ### Algebra

Simplify: (4m/5n^2)-(n/2m) A)(8m^2-5n^3)/10n^2m B)(8m^2+5n^3)/10n^2m C)(4m-n)/(5n^2-2m) D)2/5n^2 I chose C 4m/5n^2 - n/2m 4m/5(n)^2 - n/2(m) C
3. ### math

Algebra is new this year. When my problem looks like this: n(10n). I know I have to multiply 10xn, but after that do I add or multliply my answer by n?
4. ### Algebra

Add The following (8n^6 - 6n - 5n^4) + (2n^4 + 2n^6 + 3n) I think that this is the answer but am not sure. 10n^6 - 3n^4 - 3n
5. ### math

Simplify the Expressions 1) -4(3x-3) + 9(x+1)= -3x21 2) -9x(5x+4)+10(x+3)= -45x-26 x+3 3) n(1+ 10n)-n(4n-3)= 4n+6n 4) -2n(-8n-10)-6(-10n-3)= 96n - 10 5) 8(-4x-7)+3(9-9x)= 59x - 29 6) 9(3-10n)-3(10n+1)= 30n - 3 7) -4(1-8x)-9(-10x-1)= …
6. ### Physics

create a problem that could be represented/answered by the following equation. Include a sketch. Be sure to ask a question. 2kg * a = kg * 10N/kg * sin30 - 0.3 2kg * 10N/kg * cos30 so far I got that the object has a 20N downward force …
7. ### science

guyz got a hard question... A net force of 10N to a 5 kg mass. bY HOW MUCH does a body accelerate.. Given: F=10N;m=5kg
8. ### science

forces of 20N to the south,40N to the northeast, and 10N to be east act on the object.the magnitude of resultant force is a)10N b)39 N c)20 N d)46 N
9. ### Maths

In a bag there are a number of red and blue balls. The ratio of red to blue balls is 3:7. I add 10 balls and the ratio goes to 6:7 red to blue balls. How many blue balls are there?
10. ### Algebra

Write each expression in radical form. 1. (10n)^3/2 My answer: sqrt of 10n to the power of 3 Could it be simplified further to 10n times sqrt of 10n?

More Similar Questions

Post a New Question