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answer in simplest form
5c^2-45c+90(over)
180-5c^2

3x-21(over)
42-6x

6a^2-54(over)
2a^2+8a+6

simplify
6m-13(over) - 5(over)
m^2-5m+6 m-3

2007-02-20 22:16:36 · 4 answers · asked by BOBNESS 1 in Science & Mathematics Mathematics

4 answers

5c² - 45c + 90 / 180 - 5c²

- - - - - - - - -

Multiply the denominator by - 1 and rearrange

-1(180) - (- 1)(5)(c²)=

- 180 + 5c²

5c² - 180

- - - - - - - - -

Factor the numerator

5c² - 45c + 90

5(c² - 9 + 18)

5(c - 6)(c - 3)

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factor the denominator

5c² - 180

5(c² - 36)

5 (c - 6)(c + 6)

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put the numerator factors and denominator factors

5(c - 6)(c - 3) / (c - 6) ( c + 6)

The 5 and c - 6 cancell

c - 3 / c + 6

- - - - - - - -s-

2007-02-21 00:44:35 · answer #1 · answered by SAMUEL D 7 · 0 0

5c^2 - 45c + 90
-----------------
180 - 5c^2

Factor the top:
= 5(c^2 - 9c + 18)
= 5(c - 6)(c - 3)

Factor the bottom:
= 5(36 - c^2)
= 5(6 - c)(6 + c)
= -5(c - 6)(6 + c)

Now you have:
5(c - 6)(c - 3)
------------------
-5(c - 6)(6 + c)

which simplifies to:
(-1)(c - 3)
------------
(6 + c)

which simplifies to:
(3 - c)/(6 + c)
=========================
3x - 21 = 3(x - 7)
42 - 6x = 6(7 - x) = -6(x - 7)

simplifies to -1/2
====================
6a^2 - 54 = 6(a^2 - 9) = 6(a - 3)(a + 3)
2a^2 + 8a + 6 = 2(a^2 + 4a + 3) = 2(a + 3)(a + 1)

simplifies to
3(a - 3)
----------
(a + 1)
========================
I don't quite understand what you have written.

This fact should help:
when you have (a/b)/(c/d), that's the same as ad/bc

2007-02-21 06:26:50 · answer #2 · answered by Mathematica 7 · 0 0

5c^2-45c+90(over)
180-5c^2

Factor the numerator:
5(c² -9c + 18) = 5(c - 3)(c - 6).
Factor the denominator:
-5(c² - 36) = -5(c + 6)(c - 6).

Canceling common factors of 5(c - 6), the fraction simplifies to:
-(c - 3) / (c + 6).

[Approach the next two problems the same way.]

Does,
"simplify
6m-13(over) - 5(over)
m^2-5m+6 m-3"
mean,

[(6m - 13) / (-5)] / [m² + m - 3] ?

2007-02-21 06:38:18 · answer #3 · answered by S. B. 6 · 0 0

its damn hard to do it here...but it is ez. x)

2007-02-21 06:25:20 · answer #4 · answered by Xiaoboi23 1 · 0 0

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