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docker41 [41]
3 years ago
14

Find the x-intercepts of the parabola with

Mathematics
1 answer:
Llana [10]3 years ago
5 0

Answer:

The x-intercepts are (2+√(8/3),0) and (2-√(8/3),0)

Step-by-step explanation:

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The point V(-2,-1) is reflected over the y-axis. What are the coordinates of the resulting point, V’?
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(2,-1)

Step-by-step explanation:

i think it's that.

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What is the domain of the function below? {(0,2), (3,1), (5,2), (8,4)}?
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8 0
3 years ago
Which of the following is not an identity?
Shkiper50 [21]
A ) cos² x · 1 / sin x - 1 / sin x = - sin x
cos² x - 1 / sin x = - sin x 
- sin² x / sin x = - sin x
- sin x = - sin x     ( correct )
B ) sin x ( cos x / sin x + sin x / cos x ) = 1 / cos x
sin x · ( cos² x + sin²x ) / sin x cos x = 1 / cos x
sin x · 1 / sin x cos x = 1 / cos x
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C ) cos² x - sin² x = 1 - 2 sin² x
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D ) 1/sin²x + 1/ cos²x = 1
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Answer: D ) is not an identity. 
8 0
3 years ago
In a normal distribution what percent of values fall within one standard deviation of the mean?
Sophie [7]
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8 0
3 years ago
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]

Therefore, the correct option is (1,-1,2)

5 0
3 years ago
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