# Thread: Binomial expansion and exponential

1. ## Binomial expansion and exponential

1.Expand the expression using the Binomial Theorem:
________________

2.Suppose p varies inversely as the square root of q. If p=-7 when q=9, what is p if q is 6?
p=_____

3.The stopping distance d of an automobile is directly proportional to the square of its speed v. A car required 50 feet to stop when its speed was 80 miles per hour. Find a mathematical model that gives the stopping distance d in terms of its speed v.
Estimate the stopping distance if the brakes are applied when the car is traveling at 62 miles per hour.

4.If the point is on the unit circle in quadrant IV, then y=___.

For number five, let me know if these are right.

5.= sqrt(3)/2
=0
=sqrt3
= sqrt(3)/3
=-1
=2/sqrt(2)

Do you know the Binomial Theorem?
. . If you do, exactly where is your difficulty?

1. Expand the expression using the Binomial Theorem:
. . $(4x - 2)^5$

$(4x-2)^5\;=\;\binom{5}{5}(4x)^5(-2)^0 + \binom{5}{4}(4x)^5(-2)^1 + \binom{5}{3}(4x)^3(-2)^2$

. . . . . . . . $+ \binom{5}{2}(4x)^2(-2)^3 + \binom{5}{1}(4x)^1(-2x)^4 + \binom{5}{0}(4x)^0(-2)^5$

. . $= \;1(1024x^5)(1) + 5(256x^4)(-2) + 10(64x^3)(4) +$ $10(16x^2)(-8) + 5(4x)(16) + 1(1)(-32)$

. . $= \;1024x^5 - 2560x^4 + 2560x^3 - 1280x^2 + 210x - 32$

3. Oh i see what i did wrong. I didn't keep track of the negatives. Thanks. What about the others though? #2 i got -4.7 but that seems to be wrong. #3 I'm completely lost. #4 i got y=3/5 which is wrong. #5 i think is all right just want to make sure.

5.= sqrt(3)/2
=0
=sqrt3
= sqrt(3)/3
=-1
=2/sqrt(2)
I'd suggest re-checking the following:

cot(2Pi/3)

The rest look good to me.

5. Originally Posted by Soroban

Do you know the Binomial Theorem?
. . If you do, exactly where is your difficulty?

$(4x-2)^5\;=\;\binom{5}{5}(4x)^5(-2)^0 + \binom{5}{4}(4x)^5(-2)^1 + \binom{5}{3}(4x)^3(-2)^2$

. . . . . . . . $+ \binom{5}{2}(4x)^2(-2)^3 + \binom{5}{1}(4x)^1(-2x)^4 + \binom{5}{0}(4x)^0(-2)^5$

. . $= \;1(1024x^5)(1) + 5(256x^4)(-2) + 10(64x^3)(4) +$ $10(16x^2)(-8) + 5(4x)(16) + 1(1)(-32)$

. . $= \;1024x^5 - 2560x^4 + 2560x^3 - 1280x^2 + 210x - 32$

Minor error; 5(4x)(16) = 320x, not 210x