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Sub 505 Level|   Algebra|                           
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Bunuel
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can someone please explain this a bit more? I just substituted and definitely didnt come close
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can someone please explain this a bit more? I just substituted and definitely didnt come close


Hi,

the Q is --- What is the value of x^2yz − xyz^2, if x = − 2, y = 1, and z = 3? ..

\(x^2yz − xyz^2 = xyz(x-z) = (-2)*1*3*(-2-3) = (-2)*1*3*(-5) = 1*3*10=30\)...
C
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Bunuel
What is the value of x^2yz − xyz^2, if x = − 2, y = 1, and z = 3?

(A) 20
(B) 24
(C) 30
(D) 32
(E) 48


\(x^2\)\(yz\) = \(-2^2\)\((1*3)\) => 12

\(xy\)\(z^2\) = \((-2*1)3^2\) => -18

So, \(x^2\)\(yz\) − \(xy\)\(z^2\) = 12 - (-18) => 30

Hence answer will be (C) 30
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Bunuel
What is the value of x^2yz − xyz^2, if x = − 2, y = 1, and z = 3?

(A) 20
(B) 24
(C) 30
(D) 32
(E) 48


The confusion created in this question is that they could have included brackets.

ie (x^2)yz - xyz^2..

Else people like me will take x^(2yz)-xyz^2.

Is there any rule for this? Please tell me if so.
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Bunuel
What is the value of x^2yz − xyz^2, if x = − 2, y = 1, and z = 3?

(A) 20
(B) 24
(C) 30
(D) 32
(E) 48


The confusion created in this question is that they could have included brackets.

ie (x^2)yz - xyz^2..

Else people like me will take x^(2yz)-xyz^2.

Is there any rule for this? Please tell me if so.

x^2yz mathematically can only mean x^2*y*z. If it were 2 to the power of 2yz, then it would be written as x^(2yz). Still edited the original post to avoid confusion.
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Bunuel
What is the value of \(x^2yz − xyz^2\), if x = − 2, y = 1, and z = 3?

(A) 20
(B) 24
(C) 30
(D) 32
(E) 48

\((-2)^2*1*3 − (-2)*1*3^2\)

=4*1*3 - ( -2)*1*9
= 12 - (-18)
= 30

Hence answer will be (C) 30
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For this question, you must to remember that you need to factor common terms out, then it's just plug and play, and obviously, avoid silly mistakes (I'm talking directly to myself here!) and go bit by bit, just like Jack the Ripper:

\({ x }^{ 2 }yz-xy{ z }^{ 2 }=\\ xyz\left( x-y \right) =\\ \left[ \left( -2 \right) \left( 1 \right) \left( 3 \right) \right] \left( -2-3 \right) =\\ \left( -6 \right) \left( -5 \right) =\\ 30\)
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xyz(x-z)
(-2)(1)(3)(-2-3)
(-6)(-5)
30
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=> x= -2, y = 1, z = 3,

=> xyz[x-z]

=> (-2)(1)(3)(-2-3)

=> (-6)(-5)

=> 30
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\(x^2yz − xyz^2\)
\(=(-2)^2*1*3 - (-2)*1*3^2\)
\(=12+18\)
\(=30\)
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As per OG2022 the real question is something else.
Here is the image

Please help !
Attachments

WhatsApp Image 2023-02-07 at 14.26.09.jpeg
WhatsApp Image 2023-02-07 at 14.26.09.jpeg [ 80.68 KiB | Viewed 8941 times ]

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Bunuel
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PriyamRathor
As per OG2022 the real question is something else.
Here is the image

Please help !

Your OG has a typo:


Attachment:
2023-02-07_19-34-43.png
2023-02-07_19-34-43.png [ 44.66 KiB | Viewed 9238 times ]
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x^2yz − xyz^2,
= xyz (x- z)
= (-2)(1)(3) [ (-2) - 3]
= (-6)*(-5)
= 30

Posted from my mobile device
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4*1*3 -(- 2*1*9)
=12+18 = 30
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