Can someone help me with math?

1) r2-5r+6/r2-4 How would i simplify that


2) f(4) = 0 and f(6) = 6, which of the following could represent f(x) The answer is 3x-12, but i do not know y thats the answer.


3) This year, 75% of the graduating class of Harriet Tubman High School had taken at least 8 math courses. Of the remaining class members, 60% had taken 6 or 7 math


courses. What percent of the graduating class had taken fewer than 6 math courses?


4) In an arithmetic series, the terms of the series are equally spread out. For example, in


1 + 5 + 9 + 13 + 17, consecutive terms are 4 apart. If the first term in an arithmetic series is 3, the last term is 136, and the sum is 1,390, what are the first 3 terms?





5) Listed below are 5 functions, each denoted g(x) and each involving a real number constant


c %26gt; 1. If f (x) = 2x, which of these 5 functions yields the greatest value for f (g(x)), for all


x %26gt; 1 ?


A. g(x) = cx


B. g(x) = c/x


C. g(x) = x/c


D. g(x) = x – c


E. g(x) = logcx


Can u explain!

Can someone help me with math?
1.Set the equation = to 0. Solve for r with multiplication and addition/subtraction.


2.The x variable is in the ()…as in f(x) is the same as f(4)=0 then 3(4)-12=0 and f(6)=6 then 3(6)-12=6 -----%26gt; both sides must be equal when you solve. and you will see that they are equal!





hmmm here's some more to ponder:


f(4)=0


3(4)-12=0


0=0





f(6)=6


3(6)-12=6


18-12=6


6=6








3.I don’t like percentages…sorry


4.Forgot how


5.See question 2’s answer


hope this helps a little
Reply:I assume #2 is multiple choice. In that case, you have to realize that when you see f(4), you replace every x in your expression with 4. So if f(4) = 0, then 3x-12 must equal 0. 3(4) - 12 = 12 - 12 = 0, so you're good. But you probably have more than one equation for which that works, so use the second fact (f(6) = 6) to figure out which is the right answer.





#3. 75% of the class had at least 8 math courses. Of the remaining 25%, 60% of those (15% of the total) had 6 or 7. So 75% + 15% had at least 6 math courses. So 100% - 75% - 15% = 10% had fewer than 6 math courses.





#4. It's hard to explain, but I'll try. If you add the first and the last terms together, and divide the total sum by that, you'll get the number of pairs in the series. You can then multiply that by 2 to get the number of terms in the series.
Reply:Hey Bluesky. Try this for #3. If 75% took 8 courses, then 25% took some other number of courses. A piece of info. is missing, so we will assume no one took MORE than 8 classes. Thus, 25% (100%-75% = 25%) took fewer than 8 courses. Of this remaining 25%, 60% took 6 or 7 courses (60% X 25% = 15%). So, if 75% took 8, and 15% took 6 or 7, (and no one took MORE than 8) then the remaining 10% must have taken fewer than 6!





Hope you find this helpful. Best wishes and good luck.
Reply:1. (r-3)(r-2)/(r-2)(r+2)


Simplied its


(r-3)/(r+2)