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zmey [24]
3 years ago
7

Find the ratio in simplest from 30.6

Mathematics
1 answer:
kolbaska11 [484]3 years ago
8 0

Answer:

5

Step-by-step explanation:

I assume you mean "Find the ratio 30 : 6 in simplest from".

Ratio means division, so it's 30/6 = 5

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Explain why the ratio 3 feet / 1 yard has a value of one
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3 years ago
HELP WITH 4 QUESTIONS!!!!
m_a_m_a [10]

When dilation is about the origin, as it is here in every case, the image point coordinates are the original (pre-image) coordinates multiplied by the scale factor.

1. Multiply every coordinate value by 5:

... W' = (-5, 10), X' = (-15, -5), Y' = (25, -5), Z' = (15, 10)

2. Multiply every coordinate value by 1/3:

... A' = (-2, 5), B' = (0, 5/3), C' = (1, 10/3)

3. A' = (2, 8), B' = (6, 2), C' = (2, 2)

4. The image coordinates are 5 times the original coordinates, so ...

... the scale factor of the dilation is 5.

5 0
2 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 ...

\begin{bmatrix}1&3\\ 2&4\end{bmatrix} ... which is our solution.

7 0
3 years ago
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