Answer:
B(4, - 3 )
Step-by-step explanation:
We have A(2, - 7) and C(7, 3 )
Using the section formula to calculate the coordinates of B
= 
= 
= 
= 4
= 
= 
= 
= - 3
Hence B(4, - 3 )
Answer:
6 months
Step-by-step explanation:
Gym1
12 + 20m
Gym2
22m
Set them equal
12 +20m = 22m
Subtract 20 m from each side
12 +20m-20m = 22m-20m
12 = 2m
Divide by 2
12/2 = 2m/2
6 =m
Answer:
No
Step-by-step explanation:
2/3 is bigger
Answer: When t = 4.25 R(t) = D(t)
This tells you that at 4.25 seconds into their flights, both the Rocket and the Drone were at a height of 20 feet.
Step-by-step explanation: The point where the two graphed lines cross is where the values are equal.
The Drone has been hovering at 20 feet between 2 and 5 seconds. The Rocket passed through that height for a brief (millisecond) time as it was falling back to the ground.
Without grid lines it is my best estimation (using a ruler to create a verticla line from the intersection of the lines to the time scale) that t = 4.25. It is between 4 and 4.5 seconds.
Answer:
- 5.8206 cm
- 10.528 cm
- 23.056 cm^2
Step-by-step explanation:
(a) The Law of Sines can be used to find BD.
BD/sin(48°) = BD/sin(50°)
BD = (6 cm)(sin(48°)/sin(60°)) ≈ 5.82064 cm
__
(b) We can use the Law of Cosines to find AD.
AD^2 = AB^2 +BD^2 -2·AB·BD·cos(98°) . . . . . angle ABD = 48°+50°
AD^2 ≈ 110.841
AD ≈ √110.841 ≈ 10.5281 . . . cm
__
(c) The area of ∆ABD can be found using the formula ...
A = ab·sin(θ)/2 . . . . . where a=AB, b=BD, θ = 98°
A = (8 cm)(5.82064 cm)sin(98°)/2 ≈ 23.0560 cm^2
_____
Angle ABD is the external angle of ∆BCD that is the sum of the remote interior angles BCD and BDC. Hence ∠ABD = 48° +50° = 98°.
