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:
Step-by-step explanation:
The key to solving this question is to find the distance bwtween the two points given
distance=y2-y1/x2-x1
(2,4)(4,8)
2=x1,4=y1
4=x2
8=y2
8-4/4-2
4/2=2
The depth of the water is increasing by 2 ft each minute
What is being requested, if I'm not mistaken, is the number of permutations for placing each of the 8 beads on the vertices of the cubes;
In this case, we have 8 different beads and 8 possible locations for each of them;
So the number of permutations is:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320