All categories

2 years ago

PLEASE HELP, 50 POINTS! This is a triangle investigation.

Mathematics

2 answers:

mylen [45]2 years ago

5 0

A) they fornite a right angle and 2 acute angles(not 100% sure of what the Questions is asking)

B) they appear to form a triangle

C) the angles add up to 180 as they always do in all triangles

B)

Zepler [3.9K]2 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Can anyone help me pls. Show your work.

nekit [7.7K]

Answer:Do you have a pic of the map? It would make it easier to solve if I could see the map. Its basic area. A=base times height. If the park is 50 m long and is a square, you would do 50 times 50=2500

tep-by-step explanation:

4 0

3 years ago

Read 2 more answers

Write the function that shows a translation of f(x) two units to the left.

Galina-37 [17]

Answer:

Step-by-step explanation:

Suppose the base formula is y = x^2

You want to go two units left.

basically that is y = (x + 2)^2 just the opposite of what you might think it should be.

Where does the base function have a minimum? It's minimum is as x = 0 and y = 0

Where does y = (x + 2)^2 have it's minimum?

(-2,0)

conclusion. The base function has moved 2 units to the left.

3 0

3 years ago

Plz help<br>urgent !!!!<br>will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...