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alexandr402 [8]
2 years ago
14

Which shapes are similar?

Mathematics
1 answer:
shtirl [24]2 years ago
4 0

Answer:

3

Step-by-step explanation:

Since the shapes are similar this means they are equal.

If the ratio is 2:1 that would mean divide each number by 2.

6/2 reduced equals 3/1 = 3

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Which shows two expressions that are equivalent to (-8)(-12)(2)
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(-8)(-12)(2) = -8 • -12 • 2

192 //It's positive because - - = +.

Answer: 192

//Hope it helps.

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3 years ago
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the sum of two numbers is 35 think of 2 numbers that make up 35 and then subtract them and see if they make 15 {no the answer is
JulijaS [17]
25 and 10, since 25+10=35, and 25-10=15
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3^x = 27^a+b and a^2-b^2/(a-b)=5 What is x?
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Your answer should be 3x-x+2=4 hope this helps..
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Express the system as AX = B; then solve using matrix inverses
Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]

\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B

Adj.A=Transpose of cofactor of Matrix A

Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2

2.

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

   A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]

\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4

7 0
3 years ago
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