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konstantin123 [22]
2 years ago
13

I GIVE BRAINLEST !! PLS SLOLVE

Mathematics
1 answer:
alexdok [17]2 years ago
8 0

Answer:

√6/2+1

Step-by-step explanation:

multiplying by √6/√6, we get √6(3+√6)/6=(3√6+6)/6=√6/2+1

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Two lines have the equations:
butalik [34]
Let me know if you have any questions

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3 years ago
10 points)<br> (07.01 HC)<br> Prove that the two circles shown below are similar.<br> Essay question
Lapatulllka [165]

Answer:

The circles are similar.

Step-by-step explanation:

You can move the small circle 3 units to the right and then 5 units up. The center of the small circle will also be the center of the larger circle. Then you dilate the small circle by a scale factor of 5 to get the larger circle.

3 0
3 years ago
A side of the triangle has been extended to form an exterior angle of 128°, find the value of x
Tamiku [17]

Answer:

x=52

step by step:

x+128=180

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5 0
3 years ago
Carl borroowed $2,500 for six months at an annual interest rate
Margarita [4]
I think your answer is A. 
I hope I helped!
3 0
4 years ago
Please help ASAP I’ll give brainliest
butalik [34]
Write the equations in matrix,

\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\2&3&-3\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using row transformation,
R₂ <---> R₃ 

\left[\begin{array}{ccc}5&-1&1\\2&3&-3\\1&2&-1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =  \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using,
R₂ ---> R₂ - 2R₃

\left[\begin{array}{ccc}5&-1&1\\0&-1&-1\\1&2&-1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\-5\\5\end{array}\right]

Using,
R₂ --- > (-1)R₂

\left[\begin{array}{ccc}5&-1&1\\0&1&1\\1&2&-1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using row transformation,
R₂ <----> R₃
\left[\begin{array}{ccc}5&-1&1\\1&2&-1\\0&1&1\end{array}\right] \left[\begin{array}{ccc}x\\y\\z\end{array}\right] = \left[\begin{array}{ccc}4\\5\\5\end{array}\right]

Using,
R₂ ---> R₂ - R₁/5

\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&1&1\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\21/5\\5\end{array}\right]

Using,
R₃ ---> R₃ - 5R₂/11 

\left[\begin{array}{ccc}5&-1&1\\0&11/5&-6/5\\0&0&17/11\end{array}\right]   \left[\begin{array}{ccc}x\\y\\z\end{array}\right] =   \left[\begin{array}{ccc}4\\21/5\\34/11\end{array}\right]

∴ 5x-y+z = 4 ====(i)
   11y-6z = 21 === (ii)
    17z=34 === (iii)

from iii,
z=2.
Plug z=2 in ii to get y, 
∴y=3.
Plug y and z values in i to get x,
∴x=1

Therefore the solution to the system of equations is (1,3,2)
3 0
3 years ago
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