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3 years ago

What is the sum of the polynomials? (–x2 9) (–3x2 – 11x 4) –4x2 – 2x 4 –4x2 – 11x 13 2x2 20x 4 2x2 11x 5

Mathematics

1 answer:

egoroff_w [7]3 years ago

4 0

The sum of the polynomials x^2 + 9 and -3x^2 - 11x^4 - 4 is - 11x^4 -2x^2 + 5

<h3>How to determine the sum?</h3>

The polynomials are given as:

x^2 + 9

-3x^2 - 11x^4 - 4

The sum is represented as:

x^2 + 9 -3x^2 - 11x^4 - 4

Collect like terms

- 11x^4 -3x^2 + x^2 + 9 - 4

Evaluate the sum

- 11x^4 -2x^2 + 5

Hence, the sum of the polynomials is - 11x^4 -2x^2 + 5

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0.019 + (2.3 * 10-4)

DochEvi [55]

Answer:

19.019

Step-by-step explanation:

Let's solve this problem using the order of operations, or PEMDAS.

0.019 + (2.3 × 10 - 4) = 0.019 + (23 - 4) = 0.019 + 19 = 19.019

Hope this helps!

5 0

3 years ago

Read 2 more answers

Identify the standard form of the equation by completing the square.

OLEGan [10]

Answer:

$\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Step-by-step explanation:

Given equation:

$4x^2-9y^2-8x+36y-68=0$

This is an equation for a horizontal hyperbola.

To complete the square for a hyperbola

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

$\implies 4x^2-8x-9y^2+36y=68$

Factor out the coefficient of the x² term and the y² term.

$\implies 4(x^2-2x)-9(y^2-4y)=68$

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

$\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2$

$\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36$

Factor the two perfect trinomials on the left side:

$\implies 4(x-1)^2-9(y-2)^2=36$

Divide both sides by the number of the right side so the right side equals 1:

$\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}$

Simplify:

$\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1$

Therefore, this is the standard equation for a horizontal hyperbola with:

center = (1, 2)
vertices = (-2, 2) and (4, 2)
co-vertices = (1, 0) and (1, 4)
$\textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}$
$\textsf{Foci}: \quad (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)$