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3 years ago

How would you find the two triangles are similar?

Mathematics

1 answer:

harkovskaia [24]3 years ago

5 0

Answer:

The answer should be A.

Step-by-step explanation:

Well, to determine how two triangles are similar you first look at the image of what they give you.

As you can see the images shows only degree angles.

Looking at your option to see degree angles.

In some options, there is a side length of a triangle, but we aren't given a side length on both triangles so that answer does that apply.

Our last option is between A and D.

D is false as we determine the triangles are similar to each other.

RCW and VCB are vertically opposite meaning their 25° are the same to each other.

We know that every triangle adds up to 180.

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Last month, Nate worked 74 hours. This month he worked 85 hours. What percent did his hours increase? Round to the nearest hundr

Shkiper50 [21]

Percentage Change = (New - old ) / (old) *100

85 - 74 = 11

11 / 74= 0.1486 * 100 = 14.8648.

To the nearest hundredth = 14.86 %

3 0

3 years ago

I need help with this please

kodGreya [7K]

Answer:

I think it is C

Step-by-step explanation:

because x repeats with diffirent y.

7 0

4 years ago

Read 2 more answers

Question 5 (5 points)

Tpy6a [65]

Answer:

The solution of the system of equations is, (1,-1,2)

Step-by-step explanation:

Given system equation;

x + 5y - 3z = -10

-5x + 6y – 5z = -21

-x + 8y - 8z = -25

Matrix form is written as;

$\left[\begin{array}{ccc}1&5&-3\\-5&6&-5\\-1&8&-8\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}-10\\-21\\-25\end{array}\right] \\\\\\det. = 1\left[\begin{array}{cc}\\6&-5\\8&-8\end{array}\right] -5\left[\begin{array}{cc}\\-5&-5\\-1&-8\end{array}\right] -3\left[\begin{array}{cc}\\-5&6\\-1&8\end{array}\right] \\\\\\det. = 1(-8) -5(35)-3(-34)= -8 - 175+ 102 = -81$

Cofactor;

$First \ row \left[\begin{array}{cc}+\\ 6&-5\\\ 8&-8\end{array}\right \left\begin{array}{cc}-\\ -5&-5\\-1&-8\end{array}\right \left\begin{array}{cc}+\\-5&6\\-1&8\end{array}\right] = [-8 \ \ -35 \ \ -34]\\\\\\\ Second \ row \left[\begin{array}{cc}-\\ 5&-3\\\ 8&-8\end{array}\right \left\begin{array}{cc}+\\ 1&-3\\-1&-8\end{array}\right \left\begin{array}{cc}-\\1&5\\-1&8\end{array}\right] = [16\ \ -11 \ \ -13]\\\\\\$

$Third \ row \left[\begin{array}{cc}+\\ 5&-3\\\ 6&-5\end{array}\right \left\begin{array}{cc}-\\ 1&-3\\-5&-5\end{array}\right \left\begin{array}{cc}+\\1&5\\-5&6\end{array}\right]= [-7 \ \ 20\ \ 31]$

$Cofactor =$ $\left[\begin{array}{ccc}-8&-35&-34\\16&-11&-13\\-7&20&31\end{array}\right]$

$inverse \ matrix =-\frac{1}{81} \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right] \\\\\\$

Solution of the matrix:

$\left[\begin{array}{c}x\\y\\z\end{array}\right] = -\frac{1}{81} \left[\begin{array}{ccc}-8&16&-7\\-35&-11&20\\-34&-13&31\end{array}\right] X \left[\begin{array}{c}-10\\-21\\-25\end{array}\right] = \left[\begin{array}{c}\frac{-8*-10 }{-81 } +\frac{16*-21 }{-81 } + \frac{-7*-25 }{-81 }\\\\\frac{-35*-10 }{-81 } +\frac{-11*-21 }{-81 }+ \frac{20*-25 }{-81 }\\\\\frac{-34*-10 }{-81 }+ \frac{-13*-21 }{-81 }+ \frac{31*-25 }{-81 }\end{array}\right] \\\\\$

$\left[\begin{array}{c}x\\y\\z\end{array}\right] = \left[\begin{array}{c}\frac{-81}{-81} \\\\\frac{81}{-81} \\\\\frac{-162}{-81} \end{array}\right] = \left[\begin{array}{c}1\\-1\\2\end{array}\right]$