Answer:
a) Right 0≤p<π/2 and 3π/2<p≤2π
left π/2 ≤ p ≤3π/2
stopped p=π/2 and p=3π/2
b) 5.5cm and its final position (2π, 0)
c) 6.5cm
Step-by-step explanation:
we must find where the function becomes negative, positive and zero
(according to the graph)
the particle moves to the right where the function is positive
0≤p<π/2 and 3π/2<p≤2π , the particle moves to the left where the function is negative π/2 ≤ p ≤3π/2, and stopped p=π/2 and p=3π/2
s(t)=∫v(t)dt also s(t) = 5sin(t)
(according to the graph 2)
The displacement of the particle is 5.5 and its final position
(2π, 0)
The total displacement of the particle is 6.5
<em><u>Solution:</u></em>
Given that,
We have to simplify
We can simplify the above expression by combining the like terms
Like terms are terms that has same variable with same exponent and same or different coefficient
From given,
Group the like terms
Thus the given expression is simplified
Answer: the length of the angle bisector of angle ∠A is 6 ft
Step-by-step explanation:
The diagram of the right angle triangle ABC is shown in the attached photo
Looking at triangle ABC,
cos θ = adjacent side/hypotenuse side
cos θ = 5/13 = 0.3846
θ = 67.38 degrees
The bisector of an angle divides it into two equal halves. The bisector is represented by line AD.
θ/2 = 67.38/2 = 33.69 degrees
Therefore,
Cos 33.69 = 5/AD
AD × Cos 33.69 = 5
AD = 5/Cos 33.69 = 5/0.832
AD = 6 ft
Answer:
The lines representing these equations intercept at the point (-4,2) on the plane.
Step-by-step explanation:
When we want to find were both lines intercept, we are trying to find a pair of values (x,y) that belongs to both equations, which means that it satisfies both equations at the same time.
Therefore, we can use the second equation that gives us the value of y in terms of x, to substitute for y in the first equation. Then we end up with an equation with a unique unknown, for which we can solve:
Next we use this value we obtained for x (-4) in the same equation we use for substitution in order to find which y value corresponds to this:
Then we have the pair (x,y) that satisfies both equations (-4,2), which is therefore the point on the plane where both lines intercept.