Answer:
-33.5
Step-by-step explanation:
So this is a dot products vector problem, so the i and j represent x and y or horizontal and vertical components. So you multiple 3i and -0.5i and then separately multiply 8j and -4j. You get -1.5 and -32. Now all you have to do is add them and you get -33.5.
Basically in short:
1. Multiply the “i’s”.
2. Multiply the “j’s”.
3. Add the two products together.
HOPE THIS HELPS !! :-)
The arrow is fired with an initial upward velocity of 32ft/s
This question is not complete
Complete Question
A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point
Answer:
a)8.717km
b) 54.146°
Step-by-step explanation:
(a)how far is the boat from its starting point.
We solve this question using resultant vectors
= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)
Where
Rcos θ = x
Rsinθ = y
= (4cos38,4sin38) + (5cos67,5sin67)
= (3.152, 2.4626) + (1.9536, 4.6025)
= (5.1056, 7.065)
x = 5.1056
y = 7.065
Distance = √x² + y²
= √(5.1056²+ 7.065²)
= √75.98137636
= √8.7167296826
Approximately = 8.717 km
Therefore, the boat is 8.717km its starting point.
(b)calculate the bearing of the boat from its starting point.
The bearing of the boat is calculated using
tan θ = y/x
tan θ = 7.065/5.1056
θ = arc tan (7.065/5.1056)
= 54.145828196°
θ ≈ 54.146°
For the answer to the question above, the answer is simple, and it is -1 (because even powers of an imaginary number or i will always give a -1).I hope my answer helped you with your problem. Have a nice day!
Answer:
10
Step-by-step explanation:
The number of different ways that 2 lights can be chosen from 5 is ₅C₂. Use a calculator or find the value manually:
₅C₂ = 5! / (2! (5−2)!)
₅C₂ = 5! / (2! 3!)
₅C₂ = 5×4×3×2×1 / (2×1 × 3×2×1)
₅C₂ = 5×4 / 2
₅C₂ = 10
