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Vladimir79 [104]
3 years ago
7

What is the answer to M+9=2

Mathematics
2 answers:
Ainat [17]3 years ago
7 0
The answer is -7 m= -7
Ierofanga [76]3 years ago
3 0

Answer:

M= -7

Step-by-step explanation:

M+9=2

M =2-9

M=-7

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AKRAM LEFT A WEALTH OF RS 780000. HIS HEIR IS A WIDOW 3 SONS AND 4 DAUGHTERS CALCULATE THE SHARE OF EACH ONE IF THE FUNERAL EXPE
zepelin [54]

Answer:

87,500

Step-by-step explanation:

Given :

Total wealth = 780000

Number of heirs = 8

Funeral cost = 30000

Loan = 50000

Net Wealth left : (780000 - (30000 + 50000))

= 780000 - 80000

= 700000

Amount shared:

700000 / 8

= 87,500

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3 years ago
A TV that usually sells for $300 is on sale for 15% off. What is t the price of the stereo after the 15% discount. $250 $285 $25
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Answer: $255
Explanation: Multiply 300x15% to find the amount of money discounted which is $45. Then subtract the $45 from the $300 to get $255. Hope this helps!
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What is the correct transformation?
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Rotated then translated
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3 years ago
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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
I need help simplifying this equation.
mixas84 [53]

Answer:

6abc

Step-by-step explanation:

In this equation, one is given both variables and constant. One can only combine like terms; variables of the same parameter with other variables of the same parameter, or constant with constants. In this situation, three different variables are given, but two constants are given. One can combine the constant, but one must leave the variables as they are. Thus:

3 * b * 2 * ac

= 3 * 2 * b * ac

= 6abc

8 0
3 years ago
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