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1 year ago

6

which of the following number sentences illustrates the associative property of multiplication (3×8)×6=3×(8×6). 2×(1×9)=(2×1)×(2

×9). 1×15=15. 8×9=9×8.

1 answer:

Liono4ka [1.6K]1 year ago

5 0

Answer:

(A)(3×8)×6=3×(8×6)

Explanation:

Definition: A set S is said to be associative under the operation x if the following holds:

$\begin{gathered} \forall a,b,c\in S \\ a\times(b\times c)=(a\times b)\times c \end{gathered}$

Therefore, the number sentence that illustrates the associative property of multiplication is:

$\mleft(3\times8\mright)\times6=3\times\mleft(8\times6\mright).$

The correct choice is A.

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Answer:

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<u>Pre-Alg</u>

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Step-by-step explanation:

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Answer:

Second choice:

$x=2t$

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Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

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$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

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So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

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$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

$y=(x+3)^2+2(x+3)$

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

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Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

Read 2 more answers