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Klio2033 [76]
3 years ago
11

DG¯¯¯¯¯¯ , EG¯¯¯¯¯ , and FG¯¯¯¯¯ are perpendicular bisectors of the sides of △ABC . DG=5 cm and BD=12 cm.

Mathematics
1 answer:
jonny [76]3 years ago
3 0
Given that <span>DG, EG, and FG are perpendicular bisectors of the sides of △ABC, this means that the point of intersection, G, is the circumcenter of the triangle and hence A</span><span>G, BG, and CG are equal.

Given that </span><span>DG = 5 cm and BD = 12 cm, then

BG= \sqrt{DG^2+BD^2}  \\  \\ = \sqrt{5^2+12^2} = \sqrt{25+144}  \\  \\ = \sqrt{169} =13

Since AG = BG = CG, therefore, CG = 13 cm.</span>
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Answer:

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(try doing this yourself first with different points to make sure this is the right answer, because im to lazy to check myself but ill try checking D:)

Step-by-step explanation:

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so write as -2 - 3/  -7 - 2 is equal to -5/-9

so now write y = -5/9x + b

we must now find the y-intercept so substitute one of the points given. lets substitute 2,3 it looks easier. So it will look like 3 = -5/9(2) + b

3 = -5/9(2) + b

3 = -10/9 + b

b = 3 + 10/9

b = 4.111111

um just round to the tenth place

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2 years ago
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8 0
3 years ago
A bicycle depreciates at a rate of 15%. Theresa bought a bicycle for $250. How much should it be worth 6 years later?
malfutka [58]

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4 0
3 years ago
Read 2 more answers
..................................
yanalaym [24]

Answer:

The answer is a² .

Step-by-step explanation:

You have to substitute x and y into the expression :

let \: x = a \cos(θ) \\ let \: y = a \sin( θ)

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{x}^{2}  +  {y}^{2}  =  {a}^{2}( {cos}^{2}θ) +  {a}^{2} ({sin}^{2}θ)

Next, you have to apply Basic Trigonometric Identity :

{sin}^{2} θ +  {cos}^{2} θ = 1

{x}^{2}  +  {y}^{2}  =  {a}^{2} ( {cos}^{2} θ +  {sin}^{2} θ)

{x}^{2}  +  {y}^{2}  =  {a}^{2} (1)

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8 0
3 years ago
If you had carried out the algebra using variables before plugging numbers into your expressions, you would have found that (vf)
SashulF [63]

Answer:

Step-by-step explanation:

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Applying square root

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7 0
3 years ago
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