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adoni [48]
3 years ago
6

Which of the following best represents the description shown?

Mathematics
1 answer:
dolphi86 [110]3 years ago
6 0

Answer: 14+2y=78

Step-by-step explanation: the sum of something means addition so that’s why it’s gonna be 14+. Twice often means multiplication meaning 2nis gonna be multiplying with y. Equal to is an equal sign(=) meaning that y is gonna be =78. Thus explaining why 14+2y=78 is the right answer. Hope this helped:)

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87500-(3.02x10^4) / 0.00000003
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Which is the equation of the given line in point- slope form?
Alina [70]

Answer:

y-0=-1(x-8)

Step-by-step explanation:

The given line passes through (0,8) and (8,0).

The slope of this line is m=\frac{8-0}{0-8} =-1.

The point-slope form is given by:

y-y_1=m(x-x_1)

We substitute the slope and the point (8,0) to get:

y-8=-1(x-0)

We substitute the slope and the point (0,8) to get:

y-0=-1(x-8)

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Write the decimal 0.858 has a fraction.
Liula [17]

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858/1000

Step-by-step explanation:

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A survey taken of 140 sports fans asked the question: "What is the most you have ever spent for a ticket to a sporting event?" T
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The IQR is 150 - 130 = 20. Using the 1.5(IQR) criterion we get 130 -1.5(20) = 100. Since the smallest observation of 85 is smaller than 100, it should be considered an outlier

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4 years ago
If A = 1 2 1 1 and B= 0 -1 1 2 then show that (AB)^-1 = B^-1 A^-1<br><br><br> help meeeee plessss ​
Trava [24]

A = \begin{bmatrix}1&2\\1&1\end{bmatrix} \implies A^{-1} = \dfrac1{\det(A)}\begin{bmatrix}1&-1\\-2&1\end{bmatrix} = \begin{bmatrix}-1&1\\2&-1\end{bmatrix}

where det(<em>A</em>) = 1×1 - 2×1 = -1.

B = \begin{bmatrix}0&-1\\1&2\end{bmatrix} \implies B^{-1} = \dfrac1{\det(B)}\begin{bmatrix}2&1\\-1&0\end{bmatrix} = \begin{bmatrix}2&1\\-1&0\end{bmatrix}

where det(<em>B</em>) = 0×2 - (-1)×1 = 1. Then

B^{-1}A^{-1} = \begin{bmatrix}2&1\\-1&0\end{bmatrix} \begin{bmatrix}-1&1\\2&-1\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

On the other side, we have

AB = \begin{bmatrix}1&2\\1&1\end{bmatrix} \begin{bmatrix}0&-1\\1&2\end{bmatrix} = \begin{bmatrix}2&3\\1&1\end{bmatrix}

and det(<em>AB</em>) = det(<em>A</em>) det(<em>B</em>) = (-1)×1 = -1. So

(AB)^{-1} = \dfrac1{\det(AB)}\begin{bmatrix}1&-3\\-1&2\end{bmatrix} = \begin{bmatrix}-1&3\\1&-2\end{bmatrix}

and both matrices are clearly the same.

More generally, we have by definition of inverse,

(AB)(AB)^{-1} = I

where I is the identity matrix. Multiply on the left by <em>A </em>⁻¹ to get

A^{-1}(AB)(AB)^{-1} = A^{-1}I = A^{-1}

Multiplication of matrices is associative, so we can regroup terms as

(A^{-1}A)B(AB)^{-1} = A^{-1} \\\\ B(AB)^{-1} = A^{-1}

Now multiply again on the left by <em>B</em> ⁻¹ and do the same thing:

B^{-1}\left(B(AB)^{-1}\right) = (B^{-1}B)(AB)^{-1} = B^{-1}A^{-1} \\\\ (AB)^{-1} = B^{-1}A^{-1}

7 0
3 years ago
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