All categories

4 years ago

-(3/(2y+12))+2=-(8/(y+6)

Mathematics

1 answer:

musickatia [10]4 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

The perimeter of a square is given as 8x + 20 write two different expressions for the perimeter use factoring to write one of th

tatyana61 [14]

8x+20 = 4(2x+5) = 2(4x+10)

6 0

3 years ago

Please help and thank you

igor_vitrenko [27]

The answer is C) 5x^2
All you have to do is plug in the given point (1,5)

3 0

4 years ago

Evaluate the expressions. 8!-6!, 7!-5!, and 12!/8! BRAINLIST 5 min left plz hurry

nordsb [41]

Answer:

The evaluated expressions are $8!-6!=39600$ , $7!-5!=4920$ and $\frac{12!}{8!}=11880$

Step-by-step explanation:

The factorial of a positive integer n is the product of all positive integers less than or equal to n:

$n!=n\times (n-1)\times (n-2)\times (n-3)\times ...\times 3\times 2\times 1$

The factorial of n is denoted by n!

Now to find the expressions 8!-6!, 7!-5!, and 12!/8! :

Given expression is 8!-6!:

$8!-6!=[8\times (8-1)\times (8-2)\times (8-3)\times (8-4)\times (8-5)\times (8-6)\times (8-7)]- [6\times (6-1)\times (6-2)\times (6-3)\times (6-4)\times (6-5)]$ (By using factorial of n formula)

$=[8\times 7\times 6\times 5\times 4\times 3\times 2\times 1]- [6\times 5\times 4\times 3\times 2\times 1]$

$=[6\times 5\times 4\times 3\times 2\times 1][(8\times 7)-1]$

$=[720][56-1]$

$=[720][55]$

$=39600$

Therefore $8!-6!=39600$

Given expression is 7!-5! :

$7!-5!=[7\times (7-1)\times (7-2)\times (7-3)\times (7-4)\times (7-5)\times (7-6)]-[5\times (5-1)\times (5-2)\times (5-3)\times (5-4)]$ (By using factorial of n formula)

$=[7\times 6\times 5\times 4\times 3\times 2\times 1]- [5\times 4\times 3\times 2\times 1]$

$=[5\times 4\times 3\times 2\times 1][(7\times 6)-1]$

$=[120][42-1]$

$=[120][41]$

$=4920$

Therefore $7!-5!=4920$

Given expression is $\frac{12!}{8!}$

$\frac{12!}{8!}=\frac{12\times (12-1)\times (12-2)\times (12-3)\times (12-4)\times (12-5)\times (12-6)\times(12-7)\times (12-8)\times (12-9)\times (12-10) \times(12-11)}{8\times (8-1)\times (8-2)\times (8-3)\times (8-4)\times(8-5)\times (8-6)\times (8-7)}$ (By using factorial of n formula)

$=\frac{12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times4\times 3\times 2\times 1}{8\times 7\times 6\times 5\times4\times 3\times 2\times 1}$