# Need Help with complex fractions

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• Feb 25th 2007, 12:53 PM
Need Help with complex fractions

1.Match the expressions below with the letters labeling their equivalent expressions.
1.http://hosted.webwork.rochester.edu/...3a81469fc1.png

2.http://hosted.webwork.rochester.edu/...39982db4d1.png

3.http://hosted.webwork.rochester.edu/...6442b39f21.png

4.http://hosted.webwork.rochester.edu/...8f67603591.png

a.http://hosted.webwork.rochester.edu/...8215457ef1.png

b.http://hosted.webwork.rochester.edu/...32d70a06d1.png

c.http://hosted.webwork.rochester.edu/...3bf2f7e631.png

d.http://hosted.webwork.rochester.edu/...8236155ee1.png

2.Solve for a: http://hosted.webwork.rochester.edu/...0533d7c5f1.png

3.The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 998 and a standard deviation of 202 (based on date from the College Board). If a college includes a minimum score of 950 among its requirements, what percentage of females do not satisfy that requirement?

Now these I have an idea but I still need help.

These two I'm not sure weither to use P or C. Afterwards I alos need help on how to do the next steps:
4.There are nine different positions on a baseball team. If a team has 17 players how many different line-ups can the team make?
The team can make _____ different line-ups.
Baseball games consist of nine innings. A team wants to change its line-up every inning. If no game goes to extra innings, and a season consists of 190 games, how many complete seasons can the team play without repeating a line-up?
The team can play ____ complete seasons without repeating a line-up.

5.Standard automobile license plates in a country display 2 numbers, followed by 2 letters, followed by 3 numbers. How many different standard plates are possible in this system? (Assume repetitions of letters and numbers are allowed.)

Now this one I thought I had down but apparently I got it wrong. I know the area of a triangle is 1/2 base*height. I used this equation and got 184000sq.ft. When i did the work afterward the work cames out to be wrong so maybe you guys know what I did wrong.

6.A triangular parcel of land has sides of lengths 400 feet, 920 feet and 1197 feet.
a) What is the area of the parcel of land?

b) If land is valued at 1500 per acre (1 acre is 43,560 feet ), what is the value of the parcel of land?
• Feb 25th 2007, 01:17 PM
topsquark
When you have compound (complex?) fractions like you have in 1 what you need to do is clear the fractions in both the numerator and denominator.

1. You need to multiply the numerator and denominator by x^2 - 1 to clear the fraction in the numerator. Then you need to multiply the numerator and denominator by x+1 to clear the fraction in the denominator. You then get:
[x^3(x + 1)]/[x^5(x^2 - 1)] = (x + 1)/[x^2(x^2 - 1)] = (x + 1)/[x^2(x + 1)(x - 1)] = 1/[x^2(x - 1)]

See if you can get the rest. Here are what I got:
2. (x - 1)/x^2

3. x^2(x - 1)

4. x^2/(x - 1)

-Dan
• Feb 25th 2007, 01:20 PM
topsquark
Quote:

Multiply both sides by a + 5b:
a - 5b = (-2k - 4)(a + 5b)

a - 5b = -2ka - 10kb -4a - 20ba

-5b + 10kb = -2ka - 4a - 20ba - a

-5b + 10kb = (-2k - 4 - 20b - 1)a

a = (-5b + 10kb)/(-2k - 5 - 20b)

a = (5b - 10kb)/(2k + 5 + 20b)

-Dan
• Feb 25th 2007, 01:38 PM
Quote:

Originally Posted by topsquark
Multiply both sides by a + 5b:
a - 5b = (-2k - 4)(a + 5b)

a - 5b = -2ka - 10kb -4a - 20ba

-5b + 10kb = -2ka - 4a - 20ba - a

-5b + 10kb = (-2k - 4 - 20b - 1)a

a = (-5b + 10kb)/(-2k - 5 - 20b)

a = (5b - 10kb)/(2k + 5 + 20b)

-Dan

Wow thanks, I Should have got that one. I didn't even think of doing that.
• Feb 25th 2007, 01:44 PM
wait, would the answer look like this?
http://hosted.webwork.rochester.edu/...7856841951.png
• Feb 25th 2007, 02:07 PM
Jhevon
Quote:

wait, would the answer look like this?
http://hosted.webwork.rochester.edu/...7856841951.png

Actually, no. The solution looks like (-10bk - 20b)/(5 + 2k )

you guys made an error in expanding the first line. We would get -20b not a -20ba
• Feb 25th 2007, 02:27 PM
• Feb 25th 2007, 02:31 PM
Jhevon
Quote:

wait, would the answer look like this?
http://hosted.webwork.rochester.edu/...7856841951.png

Ha ha, sorry, my solution was wrong also, i left out a - 5b. Apparently this set of problems is cursed, and turns whoever tries to solve them into an idiot. I'll have the real solution in a sec
• Feb 25th 2007, 02:33 PM
Jhevon
Quote:

Originally Posted by Jhevon
Ha ha, sorry, my solution was wrong also, i left out a - 5b. Apparently this set of problems is cursed, and turns whoever tries to solve them into an idiot. I'll have the real solution in a sec

Ok, here we go. The -20b should be -15b.

So the solution is a = (-10kb - 15b)/(2k + 5)
• Feb 25th 2007, 02:35 PM
Jhevon
Quote:

The answer to number 4 is (a) x^2/(x - 1). If you want the process, tell me. I dont think you do since topsquark gave you one
• Feb 25th 2007, 02:38 PM
Jhevon
Quote:

Hold on, were you talking about #4 as in the worded problem?
• Feb 25th 2007, 02:47 PM
Jhevon
Quote:

So i'm a bit rusty in combinatorics, which is what you needed for the worded problem 4. But i'm pretty sure the answer is

17
9

= 24310 different lineups

the second answer is just that divided by 190

= 24310/190 = 127. 94 seasons without using the same line up

Note:

17
9

is to suffice for factoral notations. It means 17!/(9!(17 - 9)!)
• Feb 25th 2007, 02:59 PM
Quote:

Originally Posted by Jhevon
So i'm a bit rusty in combinatorics, which is what you needed for the worded problem 4. But i'm pretty sure the answer is

17
9

= 24310 different lineups

the second answer is just that divided by 190

= 24310/190 = 127. 94 seasons without using the same line up

Note:

17
9

is to suffice for factoral notations. It means 17!/(9!(17 - 9)!)

Yep that's what i needed thanks.
• Feb 25th 2007, 03:00 PM