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4 years ago

How would I know the domain and range? - Algebra 2

Mathematics

1 answer:

kvv77 [185]4 years ago

4 0

Domain: Set of all real numbers

Range: $-1 \le y \le 1$ where y is a real number

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The domain is the set of real numbers because there are no restrictions on what to plug in for x. There are no gaps, holes, jumps, or asymptotes. The graph continues on in both directions stretching to both infinities (this is indicated by the arrows on the curve ends)

The range is the set of y values between -1 and 1 because this is the set of y outputs. The largest y can get is +1 or just 1. The smallest y can get is -1. The value of y can be any number in between as there are no gaps. <span />

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Find the area of the triangle with the given vertices. Use the fact that the area of the triangle having u and v as adjacent sid

Gnom [1K]

Answer:

The area of the triangle is $A=\sqrt{\frac{4027}{2}}$

Step-by-step explanation:

Using the fact that the area of the triangle having u and v as adjacent sides is given by

$A=\frac{1}{2}||{\bf u} \times {\bf v} ||$

We know that we want to take a cross product to compute the area of the triangle, but we need to be careful because it doesn't make sense if we take the cross product of points.

The first step is to build some vectors that describe this triangle.

According with the graph we can build the vectors:

${\bf AB}$ and ${\bf AC}$

The vector ${\bf AB}$ is the difference of point B minus point A

${\bf AB}=(5-3,5-5,0-9)=(2,0,-9)$

and the vector ${\bf AC}$ is the difference of point C minus point A

${\bf AC}=(-4-3,0-5,2-9)=(-7,-5,-7)$

Next we need to find the cross product of this vectors.

${\bf AB} \times {\bf AC}=\begin{pmatrix}2&0&-9\end{pmatrix}\times \begin{pmatrix}-7&-5&-7\end{pmatrix}$

This is the definition of cross product of two vectors in space:

Let ${\bf u} = u_1{\bf i}+u_2{\bf j}+u_3{\bf k}$ and ${\bf v} = v_1{\bf i}+v_2{\bf j}+v_3{\bf k}$ be vectors in space. The cross product of ${\bf u}$ and ${\bf v}$ is the vector

${\bf u} \times {\bf v}=(u_2v_3-u_3v_2){\bf i}-(u_1v_3-u_3v_1){\bf j}+(u_1v_2-u_2v_1){\bf k}$

Applying this definition we get

${\bf AB} \times {\bf AC}=\begin{pmatrix}2&0&-9\end{pmatrix}\times \begin{pmatrix}-7&-5&-7\end{pmatrix}$

$\begin{pmatrix}0\cdot \left(-7\right)-\left(-9\left(-5\right)\right)&-9\left(-7\right)-2\left(-7\right)&2\left(-5\right)-0\cdot \left(-7\right)\end{pmatrix}\\\\\begin{pmatrix}-45&77&-10\end{pmatrix}$

$||{\bf AB} \times {\bf AC}||=\sqrt{(-45)^2+(77)^2+(-10)^2} \\\\||{\bf AB} \times {\bf AC}||=\sqrt{2025+5929+100}\\\\||{\bf AB} \times {\bf AC}||=\sqrt{8054}$