>>> help in mathematics <<<?

Question:(>>>help help needed in maths.<<<)
Simplify this question.

(3x^2)^3 (4y^2)^4 (3z^3)^2 devide (2x)^2 (9y^2)^3 (z)^4.

Thank you very much.
Please show detailed working step by step, in sequence so that I will understand.

Answers:
(3x^2)^3 = 27x^6
multiply the exponents 2 times 3 = 6
the coefficient is also raised to whatever the exponent is outside of the parentheses

follow that for the other terms
(4y^2)^4 = 256y^8
(3z^2)^2 = 9z^4
(2x)^2 = 4x^2
(9y^2)^3 = 729y^6
(z)^4 = z^4

now compare the exponents you have on the top and bottom of the division line:
x = 6 on top, x = 2 on bottom
because of this, you can take 2 from the x exponent on top to leave you with 4
follow this same procedure with the y and z terms
(when you get to z remember that #^0 = 1)

you should now have:
27(x^4) 256 (y^2) /divided by/ (4 times 729)

multiply your denominator numbers together
27(x^4) 256 (y^2) /divided by/ (2916)

There might be more after that... but I don't think so.
(3x^2)^3 is the same as (3 * x^2) * (3 * x^2) * (3 * x^2) =
= 27 * x^(2+2+2)= 27x^6

(4y^2)^4 is the same as (4 * y^2) * (4 * y^2) * (4 * y^2) * (4 * y^2) =
= 256 * y^(2+2+2+2) = 256y^8

(3z^3)^2 is the same as (3 * z^3) * (3 * z^3) = 9 * z^(3+3) = 9z^6

(2x)^2 is the same as (2 * x) * (2 * x) = 4 * x^(1+1) = 4x^2

(9y^2)^3 is the same as (9 * y^2) * (9 * y^2) * (9 * y^2) =
= 729 * y^(2+2+2) = 729y^6

(z)^4 is the same as z^4

so. now you right the problem again with the results that you find...

GIVEN: (3x^2)^3 (4y^2)^4 (3z^3)^2 divided by (2x)^2 (9y^2)^3 (z)^4

is the same as...

(27x^6)(256y^8)(9z^6) DIVIDED BY (4x^2)(729y^6)(z^4)

which can be written

(27x^6)(256y^8)(9z^6)
-----------------------------
(4x^2)(729y^6)(z^4)

1
X....64.....X
27x^6)(256y^8)(9z^6). 64 x^4 y^2 z^2
------------------------------... = ---------
(4x^2)(729y^6)(z^4)..3
....X
....27
.....X
.....3

64 x^4 y^2 z^2
-----------------
. 3


KEY MATH CONCEPTS YOU SHOULD KNOW:

(3y^2)^3 is the same as 3y^2 * 3y^2 * 3y^2 = 27 * y^(2+2+2) = 27y^6

or... you could have seen it this way...

(3y^2)^3 is the same as 3^3 * (y^2)^3 = 27 * y^(2*3) = 27y^6


When DIVIDING exponents... like

x^8
----
x^5

you SUBTRACT the exponents... so long as you have the same base. our base is "x"... so because the bases are the same... you can therefore subtract the exponents...

so...

x^8
---- = x^(8-5) = x^3
x^5
One variable at a time: (3x^2)^3/(2x)^2 = 27x^6/4x^2 = 27x^4/4. (For powers raised by powers, multiply them. For powers divided by powers, subtract them.) Then (4y^2)^4/(9y^2)^3 = 256y^8/729y^6 = 256y^2/729. Finally (3z^3^4/z^4) = 81z^12/z^4 = 81z^8.

The numbers on top are 27x256x81 and on the bottom are 4x729. That reduces to 3x64 = 192. The variables are x^4 x y^2 x z^8. So the final answer is 192x^4*y^2*z^8. If there are any steps you still don't get, email me.

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