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Bunuel
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For all numbers q and r, the operation Θ is defined by q Θ r = (q + 3) 03 May 2017, 04:30
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D
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91% (00:58) correct 9% (01:51) wrong based on 197 sessions

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For all numbers q and r, the operation Θ is defined by \(q Θ r = (q + 3)(r – 2)\). If \(1Θ z = –8\), then z is equal to which of the following?

(A) –3
(B) –2
(C) –1
(D) 0
(E) 2
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D
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For all numbers q and r, the operation Θ is defined by q Θ r = (q + 3) 03 May 2017, 04:58
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Given : qΘr=(q+3)(r–2)
1Θz=–8

(1+3)(z-2) = -8
z-2 = -2
x=0(Option D)
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For all numbers q and r, the operation Θ is defined by q Θ r = (q + 3) 03 May 2017, 07:11
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By simple substitution
(1+3)(Z-2)=-8
4(Z-2)=-8
z-2=-2
z=0
Option D
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For all numbers q and r, the operation Θ is defined by q Θ r = (q + 3) 06 May 2017, 17:32
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Bunuel
For all numbers q and r, the operation Θ is defined by \(q Θ r = (q + 3)(r – 2)\). If \(1Θ z = –8\), then z is equal to which of the following?

(A) –3
(B) –2
(C) –1
(D) 0
(E) 2

We are given that q@r = (q + 3)(r - 2) and that 1@z = -8.

Using the given pattern, we see that q = 1 and r = z. Thus:

1@z = (1 + 3)(z - 2)

-8 = 4(z - 2)

-2 = z - 2

0 = z

Answer: D
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For all numbers q and r, the operation Θ is defined by q Θ r = (q + 3) 29 Dec 2024, 03:12
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q Θ r = (q + 3)(r – 2) and 1 Θ z = –8

Lets start by finding the value of 1 Θ z

To find 1 Θ z we will compare what is before and after Θ in 1 Θ z and q Θ r
=> We will replace q with 1 and r with z in q Θ r = (q + 3)(r – 2) to get the value of 1 Θ z

=> 1 Θ z = (1 + 3)(z – 2) = -8 (Given)
=> 4 * (z - 2) = -8
=> z - 2 = \(\frac{-8 }{ 4}\) = -2
=> z = -2 + 2 = 0

So, Answer will be D
Hope it helps!

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