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Harrizon [31]
3 years ago
7

In the figure, segment RD bisect segment DE at S. Given that DS=4x+12 and SE=8x-8, find the value of x.

Mathematics
1 answer:
Triss [41]3 years ago
4 0

Answer:

x = 5

Step-by-step explanation:

DS = 4x + 12

SE = 8x - 8

Since segment RS bisects segment DE, it implies that the two segments created, segments DS and SE, are congruent. Therefore:

4x + 12 = 8x - 8

Solve for x

4x + 12 - 8x = 8x - 8 - 8x (Subtraction property of equality)

-4x + 12 = - 8

-4x + 12 - 12 = - 8 - 12 (subtraction property of equality)

-4x = - 20

\frac{-4x}{-4} = \frac{-20}{-4}

x = 5

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Find the outlier of the data set<br> 3.1, 5.3, 4.3, 0.7, 5.6, 5.2, 5.8, 5.4
Lelechka [254]
An outglier is a data item which is much bigger/smaller than other data items.

The data item arranged is 0.7, 3.1, 4.3, 5.2, 5.3, 5.4, 5.6, 5.8
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2 years ago
Divide the following polynomials (a4 + 4b4) ÷ (a2 - 2ab + 2b2)
telo118 [61]
(a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2+2ab+2b^2 =The answer

(a + b)^2 = a^2 + 2ab + b^2 => square of sums
(a - b)^2 = a^2 - 2ab + b^2 => square of deference
and of course one of most important ones:
a^2 - b^2 = (a - b)(a + b) => difference of squares
Best Answer: (a^4 + 4b^4) ÷ (a^2 - 2ab + 2b^2)
= [(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)] / (a^2 - 2ab + 2b^2)
= a^2 + 2ab + 2b^2
a^4 + 4b^4 => i.e. 4a^2b^2 ,
a^4 + 4a^2b^2 + 4b^4 => a^2 + 2ab + b^2 = (a + b)^2, if : a = a^2 , b = 2b^2:
(a^2 + 2b^2)^2 = a^4 + 4a^2b^2 + 4b^4 => We can't add or subtract the value to the expression.
a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 =>
(a^2 + 2b^2)^2 - 4a^2b^2 =>
(a^2 + 2b^2 - 2ab)(a^2 + 2b^2 + 2ab) =>
(a^2 - 2ab + 2b^2) (a^2 + 2ab + 2b^2)

Greetings!


8 0
3 years ago
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Answer:

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Step-by-step explanation:

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luda_lava [24]

Answer: 5/8 :)

Step-by-step explanation:

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