Ask your own question, for FREE!
Mathematics 27 Online
OpenStudy (anonymous):

write each expression so that it contains only positive exponents (2x)^-2/3y^-1

jhonyy9 (jhonyy9):

do you know the property of exponents ?

jhonyy9 (jhonyy9):

so a^-1 = ?

OpenStudy (anonymous):

1/a^1

jhonyy9 (jhonyy9):

yes so than for your exercise can using this property ?

jhonyy9 (jhonyy9):

(2x)^-2 = ?

OpenStudy (anonymous):

1/(2x)^2

OpenStudy (anonymous):

I THINK

jhonyy9 (jhonyy9):

yes right but you can calculi it again ,continue ...

OpenStudy (anonymous):

1/4X ?

jhonyy9 (jhonyy9):

why 4x ? no

OpenStudy (anonymous):

CAUSE 2*2 IS 4

jhonyy9 (jhonyy9):

yes but check it again there are (2x)^2

OpenStudy (anonymous):

2(2x) I DONT KNOW

jhonyy9 (jhonyy9):

not 2*2x this mean 2x*2x =

OpenStudy (anonymous):

4X

jhonyy9 (jhonyy9):

why ? 2x*2x = 2*2*x*x =

OpenStudy (anonymous):

I DONT KNOW HOW TO SIMPLIFY 1/(2x)^2

jhonyy9 (jhonyy9):

so (2x)^2 = 2^2 *x^2 = ?

OpenStudy (anonymous):

2*X OR 2*2X

jhonyy9 (jhonyy9):

2*2 = ?

OpenStudy (anonymous):

4

jhonyy9 (jhonyy9):

x*x=?

OpenStudy (anonymous):

MY BAD 4*X OR 4*2X

jhonyy9 (jhonyy9):

x*x= ?

OpenStudy (anonymous):

X

jhonyy9 (jhonyy9):

why ?

OpenStudy (anonymous):

IS IT 2X

jhonyy9 (jhonyy9):

so than this mean that 2*2 =2 ?

OpenStudy (anonymous):

OK

jhonyy9 (jhonyy9):

not is ok

OpenStudy (anonymous):

2*2 IS 4

jhonyy9 (jhonyy9):

yes so than x*x= ?

OpenStudy (anonymous):

ISNT X

jhonyy9 (jhonyy9):

so in your exercise there is (2x)^-2 = 1/(2x)^2 = 1/((2^2)(x^2)) = ?

OpenStudy (anonymous):

1/4(X^2)

OpenStudy (anonymous):

or 1/4(x)^2

jhonyy9 (jhonyy9):

yes but without parantheses 1/(4x^2) so this will be the first part

OpenStudy (anonymous):

ok

jhonyy9 (jhonyy9):

3y^-1 = ?

OpenStudy (anonymous):

1/3y^1

jhonyy9 (jhonyy9):

no because there is just y on the exponent -1 and 3 have not exponent yes ?

OpenStudy (anonymous):

oh

jhonyy9 (jhonyy9):

so there is 3*y^-1 = y^-1 = ?

OpenStudy (anonymous):

1/3(y)^1

jhonyy9 (jhonyy9):

no y^-1 = ?

OpenStudy (anonymous):

its 1/y^1

jhonyy9 (jhonyy9):

ok so above there is 3*y^-1 = 3*(1/y) = ?

OpenStudy (anonymous):

3y or do i do the distributive property

OpenStudy (anonymous):

3*3y distributive property way

jhonyy9 (jhonyy9):

no how you multiplie a number with a fraction ?

jhonyy9 (jhonyy9):

1 3 * ----- = ? y

OpenStudy (anonymous):

oh its 3/y

jhonyy9 (jhonyy9):

yes bravo

OpenStudy (anonymous):

lol

jhonyy9 (jhonyy9):

so after this calculi how will be your exercise ?

OpenStudy (anonymous):

good

jhonyy9 (jhonyy9):

nice

jhonyy9 (jhonyy9):

can you rewriting the first fraction ?

OpenStudy (anonymous):

you want the orginal problem

OpenStudy (anonymous):

original

OpenStudy (anonymous):

1/4x^2

jhonyy9 (jhonyy9):

1/(4x^2) / 3/y =

jhonyy9 (jhonyy9):

how you divide two fractions ?

OpenStudy (anonymous):

you want me to tell you how to divide fractions well flip the guy them multiply in this case its...

OpenStudy (anonymous):

then multiply

OpenStudy (anonymous):

so we have 1/(4x^2) / 3/y

OpenStudy (anonymous):

1/(4x^2) / y/3

jhonyy9 (jhonyy9):

so you have wrote right with words but with fractions is wrong

OpenStudy (anonymous):

dont you flip the second guy

jhonyy9 (jhonyy9):

so you write there ,,multiplie " and write divide sign next (4x^2) ,why ?

OpenStudy (anonymous):

oh forgot so its 1/(4x^2) * y/3 my bad

jhonyy9 (jhonyy9):

bravo

OpenStudy (anonymous):

:)

jhonyy9 (jhonyy9):

so the result ?

jhonyy9 (jhonyy9):

not is there the end of

OpenStudy (anonymous):

\[ 1\over 4x^2\] * Y/3

jhonyy9 (jhonyy9):

1 y ----- * ------ = ? 4x^2 3

OpenStudy (anonymous):

1Y/12X^2

jhonyy9 (jhonyy9):

yes y/(12x^2)

OpenStudy (anonymous):

SO THATS THE ANSWER

OpenStudy (anonymous):

THANK YOU VERY MUCH

jhonyy9 (jhonyy9):

yw was my pleasure good luck bye

OpenStudy (anonymous):

ARE YOU STILL THERE

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!