# Binomial expansion and exponential

• Feb 10th 2007, 01:10 PM
Binomial expansion and exponential
1.Expand the expression using the Binomial Theorem:
http://hosted.webwork.rochester.edu/...409e221a51.png___http://hosted.webwork.rochester.edu/...5f47b537b1.png__http://hosted.webwork.rochester.edu/...0e66e4f681.png___http://hosted.webwork.rochester.edu/...e25384ce81.png__http://hosted.webwork.rochester.edu/...50d06c2871.png___http://hosted.webwork.rochester.edu/...ca59068401.png___

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 http://hosted.webwork.rochester.edu/...4eff2fdd91.png is on the unit circle in quadrant IV, then y=___.

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

5.http://hosted.webwork.rochester.edu/...65f85e6fc1.png= sqrt(3)/2
http://hosted.webwork.rochester.edu/...f19bacca11.png=0
http://hosted.webwork.rochester.edu/...9f86f43a41.png=sqrt3
http://hosted.webwork.rochester.edu/...96b82a1c41.png= sqrt(3)/3
http://hosted.webwork.rochester.edu/...2720636d41.png=-1
http://hosted.webwork.rochester.edu/...d622b8eb01.png=2/sqrt(2)
• Feb 10th 2007, 02:05 PM
Soroban

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

Quote:

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

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

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

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

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

• Feb 10th 2007, 03:32 PM
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.
• Feb 10th 2007, 03:37 PM
AfterShock
Quote:

I'd suggest re-checking the following:

cot(2Pi/3)

The rest look good to me.
• Feb 10th 2007, 03:44 PM
AfterShock
Quote:

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