How would I do this
75xy^2 z^7 / 45x^4 y^3 z^6
Reducing rational expressions?
2009-03-04 3:32 pm
回答 (8)
2009-03-04 3:46 pm
✔ 最佳答案
75xy^2z^7-----------------
45x^4y^3z^6
....5z
=------------- answer//
...3x^3y
2016-10-21 8:37 pm
the reason being x^0 = a million (something to the capacity 0 is a million) least puzzling thank you to try this branch is to subtract the exponent values interior the denominator from the value interior the numerator for an identical base. We get x^(5-2)y^(3-3)z^(2-7) = x^3y^0z^-5 that's comparable to x^3/y^5
2009-03-04 3:46 pm
The original expression=
(75/45)[x^(1-4)][y^(2-3)][z^(7-6)]=
5z/(3yx^3)
(75/45)[x^(1-4)][y^(2-3)][z^(7-6)]=
5z/(3yx^3)
2009-03-04 3:40 pm
75xy²z⁷/(45x⁴y³z⁶) = 75z⁷⁻⁶/(45x⁴⁻¹y³⁻²)
= 5z/(3x³y)
= 5z/(3x³y)
2009-03-04 3:39 pm
(75xy^2z^7)/(45x^4y^3z^6)
= (75/45)(x/x^4)(y^2/y^3)(z^7/z^6)
= (5/3)(1/x^3)(1/y)(z)
= 5z/(3x^3y)
= (75/45)(x/x^4)(y^2/y^3)(z^7/z^6)
= (5/3)(1/x^3)(1/y)(z)
= 5z/(3x^3y)
2009-03-04 3:36 pm
15xy^2z^6(5z) / 15xy^2z^6(3x^3y)
=5z / 3x^3y
=5z / 3x^3y
2009-03-04 3:36 pm
5z / 3(x^3)y
2009-03-04 3:35 pm
75 on top, 45 on bottom --> 5/3
x on top, x^4 on bottom --> x^3 on the bottom
y^2 on top, y^3 on bottom --> y on the bottom
z^7 on top, z^6 on bottom --> z on top
x on top, x^4 on bottom --> x^3 on the bottom
y^2 on top, y^3 on bottom --> y on the bottom
z^7 on top, z^6 on bottom --> z on top
收錄日期: 2021-05-01 12:10:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090304073206AApDEha