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kherson [118]
2 years ago
14

If EF bisects CD, CG =5x-1, GD = 7x-13, EF = 6x-4, and GF =13, find EG

Mathematics
1 answer:
VARVARA [1.3K]2 years ago
6 0

<em>Greetings from Brasil....</em>

As stated in the statement of the question, EF is a bisector of CD, so point G is the median point of CD, so CG = GD

5X - 1 = 7X - 13

X = 6

EF = EG + GF

6X - 4 = EG + 13      x = 6, so

6·6 - 4 = EG + 13

<h2>EG = 19</h2>

<em>(see attachment)</em>

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Find the distance between the pair of points. Round your answer to the nearest hundredth. (-8,19) and (3,5)
Amiraneli [1.4K]

Answer:

<h3>The answer is 17.80 units</h3>

Step-by-step explanation:

The distance between two points can be found by using the formula

d =  \sqrt{ ({x1 - x2})^{2} +  ({y1 - y2})^{2}  } \\

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-8,19) and (3,5)

The distance between them is

d =  \sqrt{ ({ - 8 - 3})^{2}  +  ({1 9 - 5})^{2} }  \\  =  \sqrt{( { - 11})^{2}  +  {14}^{2} }  \\  =  \sqrt{121 + 196}  \\  =  \sqrt{317}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  = 17.804493...

We have the final answer as

<h3>17.80 units to the nearest hundredth</h3>

Hope this helps you

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Answer:

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Step-by-step explanation:

Simplify the following:

(-2 x^3 y^7)^2 (3 x^2 y^2)^5

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(-2)^2 x^(2×3) y^(2×7) (3 x^2 y^2)^5

2×7 = 14:

(-2)^2 x^(2×3) y^14 (3 x^2 y^2)^5

2×3 = 6:

(-2)^2 x^6 y^14 (3 x^2 y^2)^5

(-2)^2 = 4:

4 x^6 y^14 (3 x^2 y^2)^5

Multiply each exponent in 3 x^2 y^2 by 5:

4 x^6 y^14×3^5 x^(5×2) y^(5×2)

5×2 = 10:

4×3^5 x^6 y^14 x^(5×2) y^10

5×2 = 10:

4×3^5 x^6 y^14 x^10 y^10

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4×3 (3^2)^2 x^6 y^14 x^10 y^10

3^2 = 9:

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9^2 = 81:

4×3×81 x^6 y^14 x^10 y^10

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4×243 x^6 y^14 x^10 y^10

4 x^6 y^14×243 x^10 y^10 = 4 x^(6 + 10) y^(14 + 10)×243:

4×243 x^(6 + 10) y^(14 + 10)

14 + 10 = 24:

4×243 x^(6 + 10) y^24

6 + 10 = 16:

4×243 x^16 y^24

4×243 = 972:

Answer: 972 x^16 y^24

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