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Solnce55 [7]
3 years ago
9

What is the value of x? 2 3 6 7

Mathematics
2 answers:
VMariaS [17]3 years ago
7 0
This is correct it would. Be 2 and. The explanation was spot on
Arisa [49]3 years ago
5 0

Answer:

x = 2

Step-by-step explanation:

AE is equal to the sum of AB and BE because of segment addition postulate, we can substitute to get AE = 11 + x + 1 = x + 12

DE is equal to the sum of DC and CE because of segment addition postulate, we can substitute to get DE = 1 + x + 4 = x + 5

To solve the problem, we can use power of a point, more specifically exterior secants products scenario. In this case, we can get the equation:

AE * BE = DE * CE

Now, we can substitute and solve:

(x + 12)(x + 1) = (x + 5)(x + 4)

x^2 + 13x + 12 = x^2 + 9x + 20

13x + 12 = 9x + 20

4x + 12 = 20

4x = 8

x = 2

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Gre4nikov [31]

Answer:

Step-by-step explanation:

We khow that the equation of a circle is written this way :

(x-a)²+(y-b)²=r² where (x,y) are the coordinates of the cercle's points and (a,b) the coordinates of the cercle's center and r the radius .

Our task is to khow the values of a and b :

  • We khow that the center is lying on the line 3x+2y=16⇒ 2y=-3x+16⇒ y= \frac{-3}{2}x+8  
  • We khow that the points P and Q are two points in the cercle
  • Let Ω be the center of this cercle
  • we can notice that : PΩ AND QΩ are both equal to the radius ⇒ PΩ=QΩ= r
  • So let's write the expression of this distance using vectors KHOWING THAT Ω(a,b)
  • Vector PΩ(a-4,b-6) and Vector QΩ(a-8,b-2)
  • PΩ=\sqrt{(a-4)^{2}+(b-6)^{2}  } and QΩ=\sqrt{(a-8)^{2}+(b-2)^{2}   }
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