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?

- anonymous

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- hartnn

Area of rectangular (here pasture) = Length * Width

- 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

Same thing with this > Simplify. 5^–1(3^–2)
If i type this into a calculator i get a stupid number

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- hartnn

ok, treat constants and variables separately.
\(16x^7 \times 48x^4 = (16\times 48) \times(x^7 \times x^4)\)
got this ?

- anonymous

Okay... yes i think i get that... but where did the two Z come from

- hartnn

Z ?? there are no Z's there....

- anonymous

Whoops, sorry, okay, go on. I understand now

- anonymous

I think i get it
16 * 48 = 768
And 7 + 4 = 11?

- hartnn

good!

- anonymous

Okay, 1 more?

- hartnn

actually, its \(x^7 \times x^4 = x^{7+4}=x^{11}\)
and sure :)

- anonymous

Okay
something like this
Simplify (4xy2)3(xy)5
and (sorry) this
Evaluate a–4b2 for a = –2 and b = 4

- anonymous

Simplify (4xy^2)^3(xy)^5
Sorry

- hartnn

i believe the first one looks like this :
\((4xy^2)^3(xy)^5\)

- 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

okay

- anonymous

The other problem i can't figure out how to write an equation

- hartnn

\(\large Evaluate \: \: a^{–4}b^2 \:\: for\:\: a = –2 \:\:and\:\: b = 4\)
right ?

- anonymous

Here you go

##### 1 Attachment

- 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

Oh man. So confused

- anonymous

Are you doing Simplify (4xy^2)^3(xy)^5 ?

- hartnn

yes.

- anonymous

And you changed that to (4xy^2)^3=4^3x^3(y^2)^3....

- anonymous

How did you do that... that hurts my brain

- hartnn

|dw:1360891677558:dw|

- hartnn

open up your brain and try to accept new things :)

- hartnn

basically the exponent outside the bracket becomes the exponent of each of the terms inside the bracket.

- anonymous

Where did averything after that equal sign come from? Where'd the 5 go?

- anonymous

Howd 4 and the 4 more exponents come from

- 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

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

OHHHHHHHHHHHHHHHHHH

- anonymous

Yes, yes, go on (:

- 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

|dw:1360892393038:dw|

- 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

64

- hartnn

yes.
now comes the 2nd rule i posted.
\(\huge (a^m)^n=a^{mn}\)
so, what about \((y^2)^3\) ??

- anonymous

|dw:1360892627948:dw|

- hartnn

12? how ?

- anonymous

Wait

- anonymous

|dw:1360892663950:dw|

- anonymous

y^1*2

- 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

(y^2)

- hartnn

:O
the exponents are getting multiplied.
what are the 2 exponents in \((y^2)^3\)

- anonymous

Its a 3 isnt it... Sorry
(y^6)

- hartnn

yes, y^6 is correct.
so, we have now
\(64x^3y^6x^5y^5\)
lets bring 'x' terms and y terms together.

- hartnn

\(64 \: \: (x^3x^5) \:\: (y^6y^5)\)
ok, any doubts ?

- anonymous

No, i got it so far.

- 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

It would end up being 64(x^8)(y^11) ?

- anonymous

Thats the answer

- hartnn

wow! thats absolutely correct! :)

- hartnn

did you get how ?

- anonymous

Yep!

- anonymous

I get it now

- anonymous

Thank you very much!

- hartnn

welcome ^_^

Looking for something else?

Not the answer you are looking for? Search for more explanations.