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For all real numbers a, b, c, d, e, and f, the operation Θ is defined

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Joined: 04 Sep 2017
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined  [#permalink]

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20 Apr 2020, 10:30
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For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation $$(a, b, c) Θ (d, e, f) = ad + be + cf$$. What is the value of $$(1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})$$ ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02
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Re: For all real numbers a, b, c, d, e, and f, the operation Θ is defined  [#permalink]

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20 Apr 2020, 10:35
$$(1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})$$
= $$[1*1] + [(-2)*(\frac{-1}{2})] + [3*\frac{1}{3}]$$
= 1+1+1
=3
gmatt1476 wrote:
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation $$(a, b, c) Θ (d, e, f) = ad + be + cf$$. What is the value of $$(1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})$$ ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02
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Joined: 13 Apr 2020
Posts: 24
Location: United Kingdom
Schools: LBS '16 (A\$)
Re: For all real numbers a, b, c, d, e, and f, the operation Θ is defined  [#permalink]

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20 Apr 2020, 10:35
Answer = (1 x 1) + (-2 * - 1/2) + (3 * 1/3)
= 1 + 1 + 1
= 3

(E)
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Re: For all real numbers a, b, c, d, e, and f, the operation Θ is defined  [#permalink]

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20 Apr 2020, 10:36
gmatt1476 wrote:
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation $$(a, b, c) Θ (d, e, f) = ad + be + cf$$. What is the value of $$(1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})$$ ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02

As per the given conditions: $$(1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})$$

1*1=1 + (-2*-1/2=1) + (3*1/3=1) = 3 Ans E
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Re: For all real numbers a, b, c, d, e, and f, the operation Θ is defined  [#permalink]

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30 Apr 2020, 08:51
1
Top Contributor
gmatt1476 wrote:
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation $$(a, b, c) Θ (d, e, f) = ad + be + cf$$. What is the value of $$(1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})$$ ?

A. -1
B. 5/6
C. 1
D. 5/2
E. 3

PS37631.02

Given: (a, b, c) Θ (d, e, f) = ad + be + cf
So, (1,−2, 3) Θ (1, −1/2, 1/3) = (1)(1) + (-2)(-1/2) + (3)(1/3)
= 1 + 1 + 1
= 3

Cheers,
Brent
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Re: For all real numbers a, b, c, d, e, and f, the operation Θ is defined  [#permalink]

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03 May 2020, 05:15
1
gmatt1476 wrote:
For all real numbers a, b, c, d, e, and f, the operation Θ is defined by the equation $$(a, b, c) Θ (d, e, f) = ad + be + cf$$. What is the value of $$(1,−2, 3) Θ (1, −\frac{1}{2}, \frac{1}{3})$$ ?

A. -1

B. 5/6

C. 1

D. 5/2

E. 3

PS37631.02

Plugging into the equation, we have:

1(1) + (-2)(-1/2) + 3(1/3) = 1 + 1 + 1 = 3

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Re: For all real numbers a, b, c, d, e, and f, the operation Θ is defined  [#permalink]

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30 May 2020, 23:21
Given (a, b, c)Θ(d, e, f) = ad + be + cf and

$$(1,−2,3)Θ(1,\frac{−1}{2},\frac{1}{3})$$ = $$(1 * 1) + (-2 * \frac{−1}{2}) + (3 * \frac{1}{3})$$

$$(1,−2,3)Θ(1,\frac{−1}{2},\frac{1}{3})$$ = 1 + 1 + 1 = 3

Option E
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Re: For all real numbers a, b, c, d, e, and f, the operation Θ is defined   [#permalink] 30 May 2020, 23:21

For all real numbers a, b, c, d, e, and f, the operation Θ is defined

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