Ethan is proving that the slope between any two points on a straight line is the same. He has already proved that triangles 11 a
nd 22 are similar.
Drag statements and reasons to complete the proof.
Drag statements and reasons to complete the proof.
2 answers:
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Answer:
speed = 796142.197 km/hr
Step-by-step explanation:
solution
we consider here orbit around galactic center is circular so
circumference formula will be here
circumference C = 2πr ........1
here we know r = 27000
so
circumference C = 2 π × 27000
C = 169560 light year
so
we travel 169560 light year in 230000000 years
so distance will be
distance =
distance = 0.00073721 light year/year
and we know 1 year = 8760 hours
so 1 light year = 9.46027 × km
so
sped of moving is
speed = 0.00073721 × 9.46027 × km/hr
speed = 796142.197 km/hr
Answer:
0.67 mi
Step-by-step explanation:
The diagram illustrating the question is shown in the attach photo.
In triangle DCA,
Opposite = H
Adjacent = b
Angel θ = 27°
Tan θ = Opp /Adj
Tan 27° = H/b
Cross multiply
H = b x Tan 27°... (1)
From triangle DSA,
The diagram illustrating the question is shown in attach photo.
In triangle DCA,
Opposite = H
Adjacent = 2.3 – b
Angel θ = 34°
Tan θ = Opp /Adj
Tan 34° = H/ 2.3 – b
Cross multiply
H = Tan 34° (2.3 – b) .. (2)
Equating equation (1) and (2)
b x Tan 27° = Tan 34° (2.3 – b)
0.5095b = 0.6745(2.3 – b)
0.5095b = 1.55135 – 0.6745b
Collect like terms
0.5095b + 0.6745b = 1.55135
1.184b = 1.55135
Divide both side by 1.184
b = 1.55135/1.184
b = 1.31 mi
Substitute the value of b into any of the equation to obtain the height (H). In this case we shall use equation 1.
H = b x Tan 27°
H = 1.31 x Tan 27°
H = 0.67 mi
Therefore, the height of the drone is 0.67 mi
Answer: ∠JKM = and
∠JKM =
Step-by-step explanation:
Since we have given that
∠JKM=10y+6
∠MKL=8y-6
Since they are linear pairs ,
So,
∠JKM =
and
∠MKL =