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3 years ago

7

Simplify the following 16a^2b^3c^6/48ab^5c^6

1 answer:

igor_vitrenko [27]3 years ago

6 0

Answer:

a/3b^2

Step-by-step explanation:

16a^2b^3c^6/48ab^5c^6

the c^6 cancel out, then simplify a^2 and a: 16a^2b^3/48ab^5 = 16ab^3/48b^5

Then divide 48 and 16 you get 3 16ab^3/48b^5 = ab^3/3b^5

And lastly, simplify b^3 and b^5 ab^3/3b^5 =

a/3b^2

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Given the preimage and image,find the dilation scale factor.

Oliga [24]

Answer:

The dilation scale factor is $\frac{1}{2}$ .

Step-by-step explanation:

The image is the dilated form of its preimage if and only if the following conditions are observed:

1) $K' = \alpha_{1} \cdot K$

2) $T' = \alpha_{2} \cdot T$

3) $P' = \alpha_{3} \cdot P$

4) $J' = \alpha_{4} \cdot J$

5) $\alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}$

If we know that $K = (2, 0)$ , $K' = (1, 0)$ , $T = (3, 0)$ , $T' = (1.5,0)$ , $P = (1, 5)$ , $P' = (0.5, 2.5)$ , $J = (0, 3)$ and $J' = (0, 1.5)$ , then the coefficients are, respectively:

$\alpha_{1} = \frac{1}{2}$ , $\alpha_{2} = \frac{1}{2}$ , $\alpha_{3} = \frac{1}{2}$ , $\alpha_{4} = \frac{1}{2}$

As $\alpha_{1} = \alpha_{2} = \alpha_{3} = \alpha_{4}$ , we conclude that the dilation scale factor applied in the preimage is equal to $\frac{1}{2}$ .

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These four numbers are plotted on a number line: -2/3, 5/8, -3/5, and -1/2. Which is the correct ordering on the number line fro

stiks02 [169]

Answer:

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Step-by-step explanation:

Given that:

We could convert the numbers to decimal to obtain a better view of the values :

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Negative values are always arranged first on left of a number line.

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Hence,

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3 years ago