3 years ago

For which equation would n = 0 not be a solution?

1 answer:

5 0

For the first equation n + 4 = 5, because 0 + 4 is not equal 5

You might be interested in

olasank [31]

Heheheyyeyyehehhwjjwjwjjwhwhhw

8 0

4 years ago

tatyana61 [14]

$n^2+n\\\\check:\\\\a_1=1^2+1=2\\a_2=2^2+2=6\\a_3=3^2+3=12\\a_4=4^2+4=20$

6 0

3 years ago

Marizza181 [45]

With some simple rearrangement, we can rewrite the numerator as

$2x^3 - 3x^2 - x + 4 = 2(x^3 - x) - 3x^2 + x + 4 \\\\ ~~~~~~~~ = 2x(x^2-1) - 3(x^2 - 1) + x + 1 \\\\ ~~~~~~~~ = (2x-3)(x^2-1) + x+1$

Then factorizing the difference of squares, $x^2-1=(x-1)(x+1)$ , we end up with

$\dfrac{2x^3 - 3x^2 - x + 4}{x^2 - 1} = \dfrac{(2x-3)(x-1)(x+1) + x+1}{(x-1)(x+1)} \\\\ ~~~~~~~~ = \boxed{2x-3 + \dfrac1{x-1}}$

3 0

2 years ago

Eva8 [605]

Answer:

Step-by-step explanation:

3 0

3 years ago

bija089 [108]

Answer:

M=916 2/3

Step-by-step explanation:

3250-m=1500+2(m-500)

3250-M=1500+2m-1000

3250-M=2M+500

3M=2750

M=916 2/3

5 0

3 years ago