The pair of figures is similar. Find x. Round to the nearest tenth if necessary.

FIND THE AREA (LEAVE EXPLANATION)

KIM [24]

Answer:

40

Step-by-step explanation:

solve for two squares. Square 1: 32. Square 2: 8. answer: 32 + 8

3 0

3 years ago

How would I do this?

iogann1982 [59]

Answer:

A triangle is normally 180 degrees only.

Step-by-step explanation:

So, you have one 90 degree angle, and on 42.7 degree angle. Add those up to 132.7. Now, you would subtract that from 180, meaning that the answer is A, 47.3.

Good luck :DD

4 0

3 years ago

I need help is it A B C or D

nevsk [136]

The correct answer is B! ((;

8 0

3 years ago

Read 2 more answers

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

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

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

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