could someone help me on this math problem?!?

Question

(2x^7y^2/4xy^3) ^2

i totally forgot how to do this and im trying to study for my placement test that's next week. Please just tell me the steps or something! Thanks.

Answer 1

(2x^6/4y)^2
x^12/4y^2

Answer 2

Reminder: Whenever you have a power to a power, you multiply. So (x^3)^5 = x^15.

In this case, that's what you do. Since you're squaring all the terms in the brackets, all you have to do is square everything, including the powers and constants. When squaring the constants, you just square then normally. So 2 becomes 4, 7 becomes 49, and 4 becomes 16.

You might first want to reduce your answer though, as there are common factors in the numerator and denominator. You can remove an x on the top and bottom, and remove a y^2 on the top and bottom, and the 2 and 4 have a common factor of 2, too. Your new reduced fraction becomes

[(x^6)/2y]^2

In which case, you just square the terms.

(x^12)/(4y^2)

Answer 3

GIVEN:
(2x^7y^2/4xy^3)^2

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HINT:
First, raise (numerator)^2 and the (denominator)^2:

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SOLUTION:
[(2x^7y^2)^2] / [(4xy^3)^2]

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REMEMBER: When RAISING a power by a power,
it becomes more POWERFUL. . .
So we MULTIPLY the exponents together! ;o)

[(2^2)(x^14)(y^4)] / [(4^2)(x^2)(y^6)]
[(4)(x^14)(y^4)] / [(16)(x^2)(y^6)]

Now, let's reduce the leading coefficient. The number to the left of the variables(letters).
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4 and 16 are both divisible by 4. So we divide BOTH the numerator and the denominaotor by 4.

[( 4 ÷ 4 )(x^14)(y^4)] / [( 16 ÷ 4 )(x^2)(y^6)]
[(1)(x^14)(y^4)] / [(4)(x^2)(y^6)]
[(x^14)(y^4)] / [(4)(x^2)(y^6)]

Next, we REDUCE the "like" bases, the variables (the letters).
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REMEMBER: When dividing with "like" bases, we SUBTRACT the smaller exponent from the LARGER exponents to keep the exponents POSITIVE.

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(LARGE EXPONENT ─ small exponent)

[(x^14-2)(y^4-4)] / [(4)(x^2-2)(y^6-4)]
[(x^12)(y^0)] / [(4)(x^0)(y^2)]

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REMEMBER: ( STUFF )^0 = 1 unit value of stuff.
example:
2^ -1 = ½
2^0 = 1 ...........( STUFF )^0 = 1 unit value of stuff.
2^1 = 2

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SUBSTITUTE 1 for x^0 and y^0 using (parenthesis):
[(x^12)(1)] / [(4)(1)(y^2)]

Now SIMPLIFY by doing the multiplication.
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ANSWER:
(x^12) / (4y^2)

Answer 4

(2x^7y^2/4xy^3) ^2
=(2x^7y^2)/(16x^2y^6)
= x^5/8y^4

Answer 5

...=(x^6/2y)^2=x^12/4y^2.
check it for yourself, do it again, it s simple!