Yea the 2 one because 200 yea yea yea yea yea yea yea yes yea
Answer:
J(-5,2), A(-5,4), and N(-1,1)
Step-by-step explanation:
Answer:
A)82.02 mi
B) 18.7° SE
Step-by-step explanation:
From the image attached, we can see the angles and distance depicted as given in the question. Using parallel angles, we have been able to establish that the internal angle at egg island is 100°.
A) Thus, we can find the distance between the home port and forrest island using law of cosines which is that;
a² = b² + c² - 2bc Cos A
Thus, let the distance between the home port and forrest island be x.
So,
x² = 40² + 65² - 2(40 × 65)cos 100
x² = 1600 + 4225 - (2 × 2600 × -0.1736)
x² = 6727.72
x = √6727.72
x = 82.02 mi
B) To find the bearing from Forrest Island back to his home port, we will make use of law of sines which is that;
A/sinA = b/sinB = c/sinC
82.02/sin 100 = 40/sinθ
Cross multiply to get;
sinθ = (40 × sin 100)/82.02
sin θ = 0.4803
θ = sin^(-1) 0.4803
θ = 28.7°
From the diagram we can see that from parallel angles, 10° is part of the total angle θ.
Thus, the bearing from Forrest Island back to his home port is;
28.7 - 10 = 18.7° SE
I am not sure about b cut I think C is 10
Problem
For a quadratic equation function that models the height above ground of a projectile, how do you determine the maximum height, y, and time, x , when the projectile reaches the ground
Solution
We know that the x coordinate of a quadratic function is given by:
Vx= -b/2a
And the y coordinate correspond to the maximum value of y.
Then the best options are C and D but the best option is:
D) The maximum height is a y coordinate of the vertex of the quadratic function, which occurs when x = -b/2a
The projectile reaches the ground when the height is zero. The time when this occurs is the x-intercept of the zero of the function that is farthest to the right.