Question

Use the binomial theorem to expand the binomial.

(s + 4v)^5 (1 point)

here are my choices
s5 – 20s4v + 160s3v2 – 640s2v3 + 640sv4 – 1024v5
s5 + 20s4v + 160s3 + 640s2+ 1280s + 1024
s5 + 20s4v + 160s3v2 + 640s2v3 + 1280sv4 + 1024v5
s5 + 20s4v + 80s3v2 + 640s2v3 + 640sv4 + 1024v5

Answer 1

s^5 + 15s^4v + 90s^3v^2 + 270s^2v^3 + 405sv^4 + 243v^5

Answer 2

thats not a answer choice that i have

Answer 3

http://www.purplemath.com/modules/binomial2.htm

Answer 4

sorry i posted wrong 1. this link can help u

Answer 5

I posted the link in chat.

Answer 6

jenna its c

Answer 7

Tutorial: http://www.mathsisfun.com/algebra/binomial-theorem.html

Expand the following:
(s+4 v)^5
Expand (s+4 v)^5 using the binomial expansion theorem.
(s+4 v)^5 = sum_(k=0)^5 binomial(5, k) (4 v)^(5-k) s^k = binomial(5, 0) (4 v)^5 s^0+binomial(5, 1) (4 v)^4 s^1+binomial(5, 2) (4 v)^3 s^2+binomial(5, 3) (4 v)^2 s^3+binomial(5, 4) (4 v)^1 s^4+binomial(5, 5) (4 v)^0 s^5:
1024 v^5 binomial(5, 0)+256 s v^4 binomial(5, 1)+64 s^2 v^3 binomial(5, 2)+16 s^3 v^2 binomial(5, 3)+4 s^4 v binomial(5, 4)+s^5 binomial(5, 5)
Evaluate the binomial coefficients by looking at Pascal's triangle.
binomial(5, 0) = 1, binomial(5, 1) = 5, binomial(5, 2) = 10, binomial(5, 3) = 10, binomial(5, 4) = 5 and binomial(5, 5) = 1:
s^5+s^4×4 v×5+s^3 (4 v)^2 10+s^2 (4 v)^3 10+s (4 v)^4 5+(4 v)^5
Distribute exponents over products in (4 v)^2.
Multiply each exponent in 4 v by 2:
s^5+s^4×4 v×5+s^3×4^2 v^2 10+s^2 (4 v)^3 10+s (4 v)^4 5+(4 v)^5
Evaluate 4^2.
4^2 = 16:
s^5+s^4×4 v×5+s^3×16 v^2 10+s^2 (4 v)^3 10+s (4 v)^4 5+(4 v)^5
Distribute exponents over products in (4 v)^3.
Multiply each exponent in 4 v by 3:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×4^3 v^3 10+s (4 v)^4 5+(4 v)^5
In order to evaluate 4^3 express 4^3 as 4×4^2.
4^3 = 4×4^2:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×4×4^2 v^3 10+s (4 v)^4 5+(4 v)^5
Evaluate 4^2.
4^2 = 16:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×4×16 v^3 10+s (4 v)^4 5+(4 v)^5
Multiply 4 and 16 together.
4×16 = 64:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×64 v^3 10+s (4 v)^4 5+(4 v)^5
Distribute exponents over products in (4 v)^4.
Multiply each exponent in 4 v by 4:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×64 v^3×10+s×4^4 v^4 5+(4 v)^5
Compute 4^4 by repeated squaring.
4^4 = (4^2)^2:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×64 v^3×10+s (4^2)^2 v^4 5+(4 v)^5
Evaluate 4^2.
4^2 = 16:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×64 v^3×10+s×16^2 v^4 5+(4 v)^5
Evaluate 16^2.
| 1 | 6
× | 1 | 6
| 9 | 6
1 | 6 | 0
2 | 5 | 6:
s^5+s^4×4 v×5+s^3×16 v^2×10+s^2×64 v^3×10+s×256 v^4 5+(4 v)^5
Multiply 4 and 5 together.
4×5 = 20:
s^5+20 s^4 v+s^3×16 v^2×10+s^2×64 v^3×10+s×256 v^4×5+(4 v)^5
Multiply 16 and 10 together.
16×10 = 160:
s^5+20 s^4 v+160 s^3 v^2+s^2×64 v^3×10+s×256 v^4×5+(4 v)^5
Multiply 64 and 10 together.
64×10 = 640:
s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+s×256 v^4×5+(4 v)^5
Multiply 256 and 5 together.
256×5 = 1280:
s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+1280 s v^4+(4 v)^5
Distribute exponents over products in (4 v)^5.
Multiply each exponent in 4 v by 5:
s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+1280 s v^4+4^5 v^5
Compute 4^5 by repeated squaring. For example a^7 = a a^6 = a (a^3)^2 = a (a a^2)^2.
4^5 = 4×4^4 = 4 (4^2)^2:
s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+1280 s v^4+4 (4^2)^2 v^5
Evaluate 4^2.
4^2 = 16:
s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+1280 s v^4+4×16^2 v^5
Evaluate 16^2.
| 1 | 6
× | 1 | 6
| 9 | 6
1 | 6 | 0
2 | 5 | 6:
s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+1280 s v^4+4×256 v^5
Multiply 4 and 256 together.
4×256 = 1024:
Answer: |
| s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+1280 s v^4+1024 v^5

Answer 8

The answer is s^5+20 s^4 v+160 s^3 v^2+640 s^2 v^3+1280 s v^4+1024 v^5

Answer 9

FINALLY!!!!!! i have the answer and the way to work it out thank you wolfe :)

Answer 10

xD You're welcome

Answer 11

told ya it was c

Answer 12

it might take me a while to understand it all i will still be able to do so,

Answer 13

lol tester. *gives tester medal*

Answer 14

I called wolfe from chat ;)

Answer 15

@wolfe8 please help the young lady

Answer 16

If I can pull off a stunt like that again, sure.

Answer 17

lol. Tester wolfe i ready

Answer 18

ok here is the question give me a minute with the table

Which of the following equations bestrepresents the regression line for the data given in the table above?

y = x + 2
y = 2x – 2
y = –x – 2
y = x – 2

Answer 19

I want a new post and new medals. Lolz

Answer 20

this isnt the best looking table but here ya go