How the fuck i rationalize this?

How the fuck i rationalize this?

Start by getting the rest of the equation

one term at a time, my man

id start by blaming it on the patriarchy

highlight three random sentences in three random wikileaks emails to rationalize anything.

also fucking simplify it you scrub.

(b^5/2 + b^5/2)^2 all over 64b^6 = 4b^5/64b^6 = 1/16b

(2sqrt(b^5))^2/(4b^2)^3 =
4b^5/64b^6 = 1/16b

multiply by -[(4b^2)^3] on the top and bottom and simplify (enjoy the foil)

not your math army

test [math]2^2[math]

a simplified version is (sqrt(x^5)+x^(5/2))^2)/64x^6

you forgot the \ on the second math

does Sup Forums support latex anyway?

OP: what properties should i apply??

don't want the result, just wanna learn how to do it

...

1/b

Holly fuck dude... how much cost that?

like $140ish

in honors alg 2 btw

that calc does everything and is worth every penny

nice calculator bro. i'm jelly

i hope you're still taking the effort to learn the techniques right? you are doing yourself a tremendous disservice if you aren't. plus no way any uni will allow you to use that thing on an exam.

oh and underage b&

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[ b^(2/2) * b^(2/3) + \sqrt(b^5) ]^(2) / [ 4b^2 ]^3
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[ b^(5/2) + b^(5/2) ]^(2) / [ 4^(3) * b^(2*3) ]

[ 2b^(5/2) ]^(2) / [ 64 * b^(6) ]

[ 2b^(5) ] / [ 64 * b^(6) ]
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[ b^(5) ] / [ 32 * b^(6) ]
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b^(5) * 32b^(-6)
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Sup Forums

>honors algebra 2

>honors
>Algebra 2

i have to cuz he takes off in i show no work

Ah the TI N-Spire CAS. I love mine, the 3D graphing got me through calc 3.

multiply by a reciprocal
. . or I prefer to multiply by a 1

What properties did you use user?

> alg 2
> You will immediately cease and not continue to access the site if you are under the age of 18.
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Out.

k ill leave 4 chan forever, but we do this kinda shit in that

ez

>plus no way any uni will allow you to use that thing on an exam.

I used mine in all my exams. Once you get to a certain level the professors assume you can do the simple algebra.

In the Midwest honor classes are normal for everyone else.

WINRAR

THANKS user! COULD YOU PLEASE TELL ME WHAT PROPIERTIES DID YOU USE?? PLS Or atleast give me a youtube tutorial !

I'm a math major and never were we allowed to use a graphing utility on an exam. Its useless past calc 3 anyway

The terms added on the numerator are equal for starters.

b*b^(3/2) = b^(5/2)
(b^5)^(1/2) = b^(5/2)

Numerator is (2b^(5/2))^2 = 4b^5

Denominator is (4b^2)(4b^2)(4b^2) = 64b^6

4b^5 / 64b^6 = 1/16b

That took all of a 3 minutes.

multiplicative property of exponents that allows you to write a product of a base raised to different exponents as the base raised to the sum of the exponents
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I rewrote the square root part as b^(1/2) * b^5
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distributive property of exponents to simplify the denominator
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additive property of "like terms" to add the b^(5/2)'s
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4^3 = 64
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simplified the fraction by a factor of 2
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rewrote the expression as a multiplication problem to fit it neatly in one line

Enjoy your free homework for the day

You don't understand how to work with exponents. The guy who posted the picture of graph paper did it correctly but this should be obvious to you.

Look up a Khan Academy video or Patrick JMT video on adding and multiplying exponential functions.

> calculator ez mode

I'm implying the need for such a calculator in that class and that you're on Sup Forums.

After all no minors allowed

Yeah what kind of Calc 3 class lets you use a fucking CAS / graphing calculator on a test lol

Not only are you underage but a massive faggot for thinking it's cool you can't do basic algebraic factoring without a calculator.

same base, exponents sums, when the exponent is a fraction, the denominator can be moved to a root, so then we have √b^5 + √b^5 and that makes 2√b^5, and that is squared si it is 4*b^5, and then you simplify the fraction, 4b^5/64b^6 giving 1/16b as the result

Let me know if there's anything you're unsure about, OP

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Whoops, i didn't distribute the exponent in the numerator right. This user has the correct answer, kudos.
Same concepts, though

>b*b^(3/2) = b^(5/2)

I'm missing something. How does b*b^(3/2) = b^(5/2) ?

how can you be so wrong, yet so fucking sure of yourself?

Not specified user, but here's my path to this result. Sorry for shit handwriting

nice

b^1 = b^(2/2)
b^1 * b^(3/2) = b^(2/2) + b^(3/2) = b^(5/2)
2/2 + 3/2 = 5/2

and 3 minutes wasted

Also for whatever reason the image uploaded sideways, the fuck

we can picture b as b^(1) which can also be written as b^(2/2). The multiplication rule states that you add common base exponents. so b^(2/2) times b^(3/2) is the same as b^((3+2)/2) or more simply as b^(5/2)

>b*b^(3/2) = b^(5/2)

if we make the exponents equal in terms of their denominator, we can express:

b = b^(2/2) = b^1 = b

There for b^(2/2) + b^(3/2) = b^[(3+2)/2] = b^(5/2)

This would be much easier to explain with a pen and paper but I am lazy lol.

dawg u gotta FOIL the top part for a start. Don't do anything with the addition, gotta multiply first or u gonna have bad time.

>There for b^(2/2) + b^(3/2) = b^[(3+2)/2] = b^(5/2)
b^(2/2)*b^(3/2) = B^(5/2)
not in addition.

You can simplify the inside of the parentheses meaning you do not have to foil.

why b*b^(2/3) go into b^(5/2)

what app is that?

see these

simplify the last fraction then you're good to go
also, I realize that here

I would prefer to be wrong yet confident in myself and my way of thinking than not. The mistake is not the the act of being wrong, but rather the unwillingness to learn and correct oneself.

...

dubs speak the tru tru

Fuck me sorry, just replace 2/3 with 3/2
But hopefully you understand the method/reasoning

You still have to simplify
It's 1/16x

>b^(2/2) + b^(3/2)
You mean * right?
Which means you add the exponents

I don't know. If only there was a clue...

common core shill detected

Say what do you do for a job? Your handwriting seems familiar

Oops a bit of a typo

this. 1/b = rest of equation. learn2math anons.