Answer:
there is actually no pictures
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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:
Step-by-step explanation:
(47+1)/(3-6) = 48/-3 = -16
y + 1 = -16(x - 6)
y + 1 = -16x + 96
y = -16x + 95
I’m not sure but i think it is 126 pi IM SORRY IF IM WRONGGGGG
There is no such thing as a "kilowatt per hour". If that's actually what the question says, then it's a defective question, and you should put it away before it makes you any more confused.
A 120 watt light bulb uses exactly 0.12 kilowatt when it's turned on.
In one hour, it uses
(0.12 kilowatt) x (1 hour) = 0.12 kilowatt-hour of energy.
If energy costs $0.20 per kilowatt-hour, then the cost is
(0.12) x (0.20) = 2.4 cents. (0.024 dollar)
