Answer:
88
Step-by-step explanation:
Sum of angles in a quadrilateral= 360 degrees
<OCA=90degrees [ tangent perpendicular to the circle.
<OBA= 90degrees =tangent perpendicular
to the circle
<OCA + <OBA +92 + <OAB= 360
90 + 90 + 92 + <OAB = 360
272 + < OAB = 360
<0AB= 360- 272 =88 degrees
Step One
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Find the length of FO (see below)
All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)
Step Two
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Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ
Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.
FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)] \
OJ = ??
[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2 Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2 Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2 Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )
Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.
Answer:
The answer is option A) X + 2
Step-by-step explanation:
Any polynomial of order n can be expressed in terms of their roots in factors of the form
(x-a)(x-b)(x-c)(x-d)....... = 0
Where a,b,c,d,... are the roots of the polynomial
In the example shown
(x+2) = (x-(-2))
And -2 happens to be the root of the polynomial in question
