anonymous
  • anonymous
A rectangular pasture has a fence around the perimeter. The length of the fence is 16x7 and the width is 48x4. What is the area of the pasture? (1 point) a) 3x^3 b)128x^11 c)768x^11 d)768x^28 I KNOW the answer is c, but i need it explained. I already found it here once but can ANYONE explain how to do this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hartnn
  • hartnn
Area of rectangular (here pasture) = Length * Width
anonymous
  • anonymous
Allright, so how do i do it? 16 * ^7 gives nothing and im so confused How do i do 16x^7 * 48x^4? I dont get this problem
anonymous
  • anonymous
Same thing with this > Simplify. 5^–1(3^–2) If i type this into a calculator i get a stupid number

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hartnn
  • hartnn
ok, treat constants and variables separately. \(16x^7 \times 48x^4 = (16\times 48) \times(x^7 \times x^4)\) got this ?
anonymous
  • anonymous
Okay... yes i think i get that... but where did the two Z come from
hartnn
  • hartnn
Z ?? there are no Z's there....
anonymous
  • anonymous
Whoops, sorry, okay, go on. I understand now
anonymous
  • anonymous
I think i get it 16 * 48 = 768 And 7 + 4 = 11?
hartnn
  • hartnn
good!
anonymous
  • anonymous
Okay, 1 more?
hartnn
  • hartnn
actually, its \(x^7 \times x^4 = x^{7+4}=x^{11}\) and sure :)
anonymous
  • anonymous
Okay something like this Simplify (4xy2)3(xy)5 and (sorry) this Evaluate a–4b2 for a = –2 and b = 4
anonymous
  • anonymous
Simplify (4xy^2)^3(xy)^5 Sorry
hartnn
  • hartnn
i believe the first one looks like this : \((4xy^2)^3(xy)^5\)
hartnn
  • hartnn
ok, so you need to remember/understand few thing before u start: \(\huge (ab)^n = a^nb^n\) and \(\huge (a^m)^n=a^{mn}\)
anonymous
  • anonymous
okay
anonymous
  • anonymous
The other problem i can't figure out how to write an equation
hartnn
  • hartnn
\(\large Evaluate \: \: a^{–4}b^2 \:\: for\:\: a = –2 \:\:and\:\: b = 4\) right ?
anonymous
  • anonymous
Here you go
1 Attachment
hartnn
  • hartnn
ok, for the previous one : using, \(\huge (ab)^n = a^nb^n\) \(\large (4xy^2)^3= 4^3 x^3 (y^2)^3\) got this ?
anonymous
  • anonymous
Oh man. So confused
anonymous
  • anonymous
Are you doing Simplify (4xy^2)^3(xy)^5 ?
hartnn
  • hartnn
yes.
anonymous
  • anonymous
And you changed that to (4xy^2)^3=4^3x^3(y^2)^3....
anonymous
  • anonymous
How did you do that... that hurts my brain
hartnn
  • hartnn
|dw:1360891677558:dw|
hartnn
  • hartnn
open up your brain and try to accept new things :)
hartnn
  • hartnn
basically the exponent outside the bracket becomes the exponent of each of the terms inside the bracket.
anonymous
  • anonymous
Where did averything after that equal sign come from? Where'd the 5 go?
anonymous
  • anonymous
Howd 4 and the 4 more exponents come from
hartnn
  • hartnn
(4xy^2)^3(xy)^5 <------you are asking about this 5 ? i am starting with (4xy^2)^3 part of (4xy^2)^3(xy)^5
hartnn
  • hartnn
if you are asking from where does all the exponents come from take a look at this again : \((ab)^n=a^nb^n\) or perhaps : \((abc)^n=a^nb^nc^n\)
anonymous
  • anonymous
OHHHHHHHHHHHHHHHHHH
anonymous
  • anonymous
Yes, yes, go on (:
hartnn
  • hartnn
so i assume you are clear with this diagram.|dw:1360892207194:dw| using same rule, can you tell me what u get for \((xy)^5\) ??
anonymous
  • anonymous
|dw:1360892393038:dw|
hartnn
  • hartnn
correct :) see, it isn't difficult at all. now you have \(\large 4^3x^3(y^2)^3x^5y^5\) right ? lets start simplifying with constants (numbers) 4^3 =... ?
anonymous
  • anonymous
64
hartnn
  • hartnn
yes. now comes the 2nd rule i posted. \(\huge (a^m)^n=a^{mn}\) so, what about \((y^2)^3\) ??
anonymous
  • anonymous
|dw:1360892627948:dw|
hartnn
  • hartnn
12? how ?
anonymous
  • anonymous
Wait
anonymous
  • anonymous
|dw:1360892663950:dw|
anonymous
  • anonymous
y^1*2
hartnn
  • hartnn
umm..no, let me give you an example \(\large (z^5)^3 = z^{5 \times 3 }=z^{15}\) do similar thing for \((y^2)^3 \)
anonymous
  • anonymous
(y^2)
hartnn
  • hartnn
:O the exponents are getting multiplied. what are the 2 exponents in \((y^2)^3\)
anonymous
  • anonymous
Its a 3 isnt it... Sorry (y^6)
hartnn
  • hartnn
yes, y^6 is correct. so, we have now \(64x^3y^6x^5y^5\) lets bring 'x' terms and y terms together.
hartnn
  • hartnn
\(64 \: \: (x^3x^5) \:\: (y^6y^5)\) ok, any doubts ?
anonymous
  • anonymous
No, i got it so far.
hartnn
  • hartnn
now one of the very important rule : \(\large a^m a^n = a^{m+n}\) here, if we multiply the variables, their exponents gets ADDED . so, what about \(x^3x^5=... ?\)
anonymous
  • anonymous
It would end up being 64(x^8)(y^11) ?
anonymous
  • anonymous
Thats the answer
hartnn
  • hartnn
wow! thats absolutely correct! :)
hartnn
  • hartnn
did you get how ?
anonymous
  • anonymous
Yep!
anonymous
  • anonymous
I get it now
anonymous
  • anonymous
Thank you very much!
hartnn
  • hartnn
welcome ^_^

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