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4 years ago

What conclusions can you draw from the distances that you recorded in question 7

Mathematics

2 answers:

Lubov Fominskaja [6]4 years ago

8 0

Answer:

If corresponding vertices on an image and a preimage are connected with line segments, the line segments are divided equally by the line of reflection. That is, the perpendicular distance from the line of reflection to either of the corresponding vertices is the same. Line EF is a perpendicular bisector of the connecting line segments.

Step-by-step explanation:

PLATO

tangare [24]4 years ago

6 0

Answer:

If corresponding vertices on an image and a preimage are connected with line segments, the line segments are divided equally by the line of reflection. That is, the perpendicular distance from the line of reflection to either of the corresponding vertices is the same. Line is a perpendicular bisector of the connecting line segments.

Step-by-step explanation:

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Please help me with geometry i really need help

Debora [2.8K]

The ratio is 1:2 meaning you would divide the 4320 by 2 to get the smaller prisim.
4320 ÷ 2 = 2160 cm^3

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3 years ago

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Rs = 8y + 4 , ST = 4y + 8 , and RT = 36 , find the value of y

Natasha_Volkova [10]

I assume that you meant RS and ST are segments of RT. If that is true then:

RS+ST=RT, using the values for these given...

8y+4+4y+8=36 combine like terms on left side

12y+12=36 subtract 12 from both sides

12y=24 divide both sides by 12

y=2

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3 years ago

What is the sum of (2+4i) and (7+11i)

Romashka [77]

Add like terms
2+4i +7 +11 i =
2+7 + 4i+11i =
9+15i

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4 years ago

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Johny has 100 bottels of dish soap

sveta [45]

Johnny now has 90 so Jeana did not take 25 she only took 10

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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 ...

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

7 0

4 years ago