For example, the additive inverse of 12 is –12. The additive inverse of –3 is 3. Formally, the additive inverse of x is –x. Note: The sum of a number and its additive inverse is 0.

8 0

4 years ago

The sum of the first three terms of a sequence is 6 and the fourth term is 16

Sati [7]

Answer:

a₁ = -5, d = 7, a₂ = 2, a₃ = 9, a₄ = 16

equation of sequence: $\boxed{\bold{a_n=7n-12}}$

Step-by-step explanation:

a₁ + a₂ + a₃ = 6

a₁ + a₁ + d + a₁ + 2d = 6

3a₁ + 3d = 6

a₁ + d= 2 ⇒ a₁ = 2 - d

a₄ = 16

a₁ + 3d = 16

2 - r + 3d = 16

2d = 14

d = 7

a₁ = 2-7 = -5

a₁ = -5, d = 7 ⇒ a₂ = -5+7 = 2, a₃ = 2+7 = 9, a₄ = 9+7 = 16

equation of arithmetic sequence:

$a_n=a_1+d(n-1)\\\\a_n=-5+7(n-1)\\\\\underline{a_n=7n-12}$

6 0

3 years ago

Look at picture please worth 15 points please helpppp

sergij07 [2.7K]

Answer:

3x - y + $\frac{- 5x^2y}{3x^3 + 2x^2y - xy^2}$

Step-by-step explanation:

Hooboy okay here we go.

1.) Rewrite the equation in the classic division method

3x³ + 2x²y - xy² $\sqrt{9x^4 + 3x^3y - 5x^2y^2 + xy^3}$

2.) Divide using normal division rules, but it's super hard now 'cause there are letters involved. Basically, what can you multiply 3x³ by to get 9x⁴? 3x. So 3x is your first number. Then multiply everything else by 3x, and subtract

3x³ + 2x²y - xy² $\sqrt{9x^4 + 3x^3y - 5x^2y^2 + xy^3}$

-(9x⁴ + 6x³y - 3x²y²)

= 0 - 3x³y - 2x²y² - 5x²y + xy³

Now lets do that again, see what happens. 3x³ times -y should get -3x³y

3x - y

3x³ + 2x²y - xy² $\sqrt{9x^4 + 3x^3y - 5x^2y^2 + xy^3}$

- (9x⁴ + 6x³y - 3x²y²)

= 0 - 3x³y - 2x²y² - 5x²y + xy³

- ( - 3x³y - 2x²y² + xy³)

= 0 + 0 - 5x²y + 0

So! From this information, we can know that the answer is 3x-y with a remainder of - 5x²y. How it's written is shown above, in "Answer"

3 0

3 years ago