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3 years ago

Factor p^2-q^2-2p+2q

Mathematics

2 answers:

Grace [21]3 years ago

6 0

Not sure of the answer but i know what factors
p ^2 - 2p

p (p-2)

-q^2 +2q

-q (q-2)

so maybe (p-q) (p-2) (q-2)

Genrish500 [490]3 years ago

4 0

Not sure of the answer but i know what factors
p ^2 - 2p

p (p-2)

-q^2 +2q

-q (q-2)

so maybe (p-q) (p-2) (q-2)

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

4 0

2 years ago

Which two (2) points on the number line below are opposites?

dexar [7]

Answer:

B. B and E

Step-by-step explanation:

A: -5.5

B: -4

C: -2.5

D: 1.5

E: 4

6 0

2 years ago

Read 2 more answers

Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

ivanzaharov [21]

Answer:

$c = \displaystyle\frac{9}{4}$

Step-by-step explanation:

The following information is missing in the given question:

$f(x) = \sqrt{x}$

Using this we may solve the question as:

We are given the following in the question:

$f(x) = \sqrt{x}, x \in [0,9]$

We have to find the number c such that f(x) satisfies the Mean value theorem.

Mean Value theorem:

It states that if the function is differentiable in the closed interval [a,b], differentiable in the interval (a,b), then there exist c in (a,b) such that:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}$

Now,

$f(x) = \sqrt{x}\\f'(x) = \frac{1}{2\sqrt{x}}$

Continuity in [0,9]

Since a polynomial function is continuous everywhere, f(x) is continuous in [0,9]

Differentiability in (0,9)

Since a polynomial function is differentiable everywhere the given function is differentiable in interval (0,9)

Then, by mean value theorem:

$f'(c) = \displaystyle\frac{f(b)-f(a)}{b-a}\\\\\frac{1}{2\sqrt{c}} = \frac{f(9) - f(0)}{9-0} = \frac{3}{9}\\\\\frac{1}{2\sqrt{c}} = \frac{1}{3}\\\\c= \frac{9}{4}$