Match the products of rational expressions with their simplest forms.

Mathematics

1 answer:

Alborosie2 years ago

7 0

Lets simplify each one of the rational expressions using the product rule of exponents: $a^n*a^m=a^{n+m}$ and the quotient rule of exponents: $\frac{a^n}{a^m} =a^{n-m}$ .

1. $\frac{-3x^2}{2y^2} * \frac{y^2}{9x} =- \frac{x}{6} = \frac{-1}{6} x$

2. $\frac{5y^2}{10x^2} * \frac{4x^2}{y} =2y$

3. $\frac{-9y^2}{5x} * \frac{-10x^2}{3y} =6xy$

4. $\frac{14x^2}{5y} * \frac{-10y}{7x} =-4x$

We can conclude that you should pair the rational expressions with their simplest form as follows:
$6xy-----\ \textgreater \ \frac{-9y^2}{5x} * \frac{-10x^2}{3y}$
$\frac{-1}{6} x-----\ \textgreater \ \frac{-3x^2}{2y^2} * \frac{y^2}{9x}$
$-4x-----\ \textgreater \ \frac{14x^2}{5y} * \frac{-10y}{7x}$
$2y-----\ \textgreater \ \frac{5y^2}{10x^2} * \frac{4x^2}{y}$

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Y=x^2-6x-16 in vertex form

satela [25.4K]

Answer:

$y=(x-3)^{2} -25$

Step-by-step explanation:

The standard form of a quadratic equation is $y=ax^{2} +bx+c$

The vertex form of a quadratic equation is $y=a(x-h)^{2} +k$

The vertex of a quadratic is (h,k) which is the maximum or minimum of a quadratic equation. To find the vertex of a quadratic, you can either graph the function and find the vertex, or you can find it algebraically.

To find the h-value of the vertex, you use the following equation:

$h=\frac{-b}{2a}$

In this case, our quadratic equation is $y=x^{2} -6x-16$ . Our a-value is 1, our b-value is -6, and our c-value is -16. We will only be using the a and b values. To find the h-value, we will plug in these values into the equation shown below.

$h=\frac{-b}{2a}$ ⇒ $h=\frac{-(-6)}{2(1)}=\frac{6}{2} =3$

Now, that we found our h-value, we need to find our k-value. To find the k-value, you plug in the h-value we found into the given quadratic equation which in this case is $y=x^{2} -6x-16$

$y=x^{2} -6x-16$ ⇒ $y=(3)^{2} -6(3)-16$ ⇒ $y=9-18-16$ ⇒ $y=-25$