## Tony.M Group Title (b^4)^6 * (b^2)^4 Simplify ? i need help one year ago one year ago

1. kirbykirby Group Title

(b^4)^6=b^24 (b^2)^4=b^8 b^24*b^8 = b^32 When you have the form (a^x)^y... u multiply the exponents so a^(xy) When you have the form b^x*b^y, you add the exponents so b^(x+y)

2. Tony.M Group Title

ok so does the formula stay the same if the problem read (k^3d^5)^5 ? im getting a bit confused on this one

3. kirbykirby Group Title

yes you "distribute" the ^5... so you get k^15 * d^25

4. kirbykirby Group Title

you can't simplify further since you have 2 different variables

5. Tony.M Group Title

thanks so much :) could you help me with a area and dimensions problem?...i dont want the answer i just want to know how to set up the problem

6. kirbykirby Group Title

I could try my best :)

7. Tony.M Group Title

Justin wants to use 188 ft of fencing to fence off the greatest possible rectangular area for a garden. how would i find what dimensions he should use and what the area would be

8. kirbykirby Group Title

Is this a calculus question

9. Tony.M Group Title

yea i think so

10. kirbykirby Group Title

Ok so you want to maximize the area of the rectangle. The area of a rectangle is A=b*L (area=base*length). We know the perimeter of a rectangle is 2(b+L)=188. Since you want to maximize A, you want to express A with only one variable. So, you use the perimeter equation above and solve, for say L, and substitute the expression for L into your Area formula. So now you have an equation A = ... (which only has one variable "b"). So, know you just differentiate your function and setting it equal to 0 to find the critical points. Make sure you only consider actual physical values (if you get a negative value, for example, ignore it since you can't have a negative length). So, then you evaluate your function A = ... at the critical points and determine which value gives the largest value for A. This will be your maximum area.

11. kirbykirby Group Title

I hope that's clear :S

12. kirbykirby Group Title

It's funny you ask this question because I have a review assignment for another class based on first year calculus and the problem I am attacking now is an optimization problem as well :P

13. kirbykirby Group Title

Oh wait I partially answered the question... You are asking for the dimensions. So, the process I told you will let you find "b". So with that, you can plug is the value for b in the perimeter equation 2(b+L)=188 and solve for L to get the length

14. Tony.M Group Title

its very clear just hard to understand because i have ADD which makes it kind of hard to read everything without forgetting -_- im going to test it and post my answer can you tell me if im right?

15. kirbykirby Group Title

Ok sure

16. Tony.M Group Title

i got 8,811 ft^2 ?

17. kirbykirby Group Title

Um I got 2209

18. Tony.M Group Title

:/ i went wrong somewhere

19. kirbykirby Group Title

If you think of it in a simple way, although this depends if your prof talked about it or not, but the greatest area rectangle you can get is always a square. So 188 = 4c (c = side of a square) so c = 47. So each side is 47 ft. The area of a square = c^2 = 47^2 = 2209

20. kirbykirby Group Title

Do you know how u got 8811?

21. Tony.M Group Title

thats what i just got 47X47 47; 2,209 ft2 :) i went back over your steps..maths so hard !

22. kirbykirby Group Title

Yay awesome :)!

23. Tony.M Group Title

one more favor ...can you help me simplify an expression? i pretty much got the rest

24. kirbykirby Group Title

Ok sure

25. Tony.M Group Title

4x^2y^7 over 24x^3y^4

26. kirbykirby Group Title

$\frac{ 4x^2y^7 }{ 24x^3y^4 }=\frac{ 1x^{2-3}*y^{7-4} }{ 6 }=\frac{ x^{-1}*y^{3} }{ 6 }=\frac{ y^{3} }{ 6x }$

27. Tony.M Group Title

THANKS SO MUCH! wish i could give you a million medals :)

28. kirbykirby Group Title

hm ok the numbers look small but it says x^(2-3) in the second "="

29. kirbykirby Group Title

:) No problem!