Help me Solve 5^-3 please

HELP PLEASE !!! Place each expression under the equivalent expression in the table.

tino4ka555 [31]

Answer:

Option 3,5 are under first expression and Option 1,2 are under second expression.

Step-by-step explanation:

1). $\frac{(x-4)(x+2)}{x^{2}+5x+6}+\frac{-3x^{2}+24x-20}{(x+3)(4x-5)}$

$=\frac{(x-4)(x+2)}{(x+3)(x+2)}+\frac{-3x^{2}+24x-20}{(x+3)(4x-5)}$ $=\frac{(x-4)}{(x+3)}+\frac{-3x^{2}+24x-20}{(x+3)(4x-5)}$

$=\frac{(x-4)(4x-5)-3x^{2}+24x-20}{(x+3)(4x-5)}$

$=\frac{4x^{2}-5x-16x+20-3x^{2}+24x-20}{(x+3)(4x-5)}$ $=\frac{x^{2}+3x}{(x+3)(4x-5)}$

$=\frac{x(x+3)}{(x+3)(4x-5)}=\frac{x}{4x-5}$

2). $\frac{3x}{4x-5}-\frac{4x^{2}}{8x^{2}-10x}=\frac{3x}{4x-5}-\frac{4x}{8x-10}$ $=\frac{3x(8x-10)-4x(4x-5)}{(4x-5)(8x-10)}$

$=\frac{24x^{2}-30x-16x^{2}+20x}{(4x-5)(8x-10)}$

$=\frac{8x^{2}-10x}{(4x-5)(8x-10)}$ $=\frac{x(8x-10)}{(4x-5)(8x-10)}=\frac{x}{4x-5}$

3). $\frac{6x^{2}}{x^{2}-7x+10}\div \frac{2x}{x-5}$

$=\frac{6x^{2}}{x^{2}-7x+10}\times \frac{x-5}{2x}$

$=\frac{3x(x-5)}{x^{2}-7x+10}$

$=\frac{3x(x-5)}{(x-5)(x-2)}=\frac{3x}{x-2}$

4). $\frac{-x}{4x-5}-\frac{4x^{2}}{16x^{2}-22x}$

$=\frac{-x}{4x-5}-\frac{2x}{8x-11}$

$=\frac{-x(8x-11)-2x(4x-5)}{(4x-5)(8x-11)}$

$=\frac{-8x^{2}+11x-8x^{2}+10x}{(4x-5)(8x-11)}$

$=\frac{-16x^{2}+21x}{(4x-5)(8x-11)}=\frac{-x(16x-21)}{(4x-5)(8x-11)}$

5). $\frac{3x^{2}}{x+3}\times \frac{2(x+3)}{2x^{2}-4x}$

$=\frac{6x^{2}(x+3)}{2x(x+3)(x-2)}=\frac{3x}{x-2}$

6). $\frac{5x^{2}}{x-2}\times \frac{2x+6}{8x^{2}-4x}$

$=\frac{10x^{2}(x+3)}{4x(x-2)(2x-1)}$

4 0

3 years ago

Read 2 more answers

6.849 to the nearest hundredth

diamong [38]

Answer:

6.85

Step-by-step explanation:

5 0

3 years ago

Question 20 read the question and type your response in the box below the question. your response will be saved automatically. w

grandymaker [24]

You're looking for something like y=2x+3 and 2y=4x+6?

3 0

3 years ago

The chart below represents the number of marbles in a jar. P(green)

Dmitry_Shevchenko [17]

Answer:

a

Step-by-step explanation:

you have to add all the numbers then put the number of marbles its asking for over it . if you can simplify then do that too, but in this case you cant .

3 0

3 years ago

the number of people who own computers has increased by 23.2 percent annually since 1990. if half a million people owned a compu

kap26 [50]

There would be at least 92 million people who will own a computer in 2015.

I don't know how to make a function out of a percentage, that is why I used excel to compute annual population.

YEAR POPULATION Increase of previous years population
1990 500,000
 1991 616,000 1.232
 1992 758,912 1.232
 1993 934,978 1.232
 1994 1,151,894 1.232
 1995 1,419,134 1.232
 1996 1,748,373 1.232
 1997 2,153,996 1.232
 1998 2,653,723 1.232
 1999 3,269,387 1.232
 2000 4,027,885 1.232
 2001 4,962,354 1.232
 2002 6,113,620 1.232
 2003 7,531,980 1.232
 2004 9,279,400 1.232
 2005 11,432,221 1.232
 2006 14,084,496 1.232
 2007 17,352,099 1.232
 2008 21,377,786 1.232
 2009 26,337,433 1.232
 2010 32,447,718 1.232
 2011 39,975,588 1.232
 2012 49,249,925 1.232
 2013 60,675,908 1.232
 2014 74,752,718 1.232
 2015 92,095,349

7 0

3 years ago