It is roughly 0.707 so it should be very close to .7
(0,-8)
Explanation: In y=mx + b, m represents the slope of the line and b represents the y intercept, since b is -8, it will cross the y-axis at (0,-8)
Answer:
θ = 5π/6 rad and 11π/6 rad
Step-by-step explanation:
Given the expression cotθ+√3=0
Subtract √3 from both sides
cotθ+√3-√3=0-√3
cotθ = -√3
Since cotθ = 1/tanθ
1/tanθ = -√3
Reciprocate both sides:
tanθ = -1/√3
θ = tan^-1(-1/√3)
θ = -30°
Since the angle is negative, and tanθ is negative in the second and fourth quadrant.
In the second quadrant;
θ = 180-30
θ = 150°
Since 180° = πrad
150° = 150π/180
150° = 5π/6 rad
In the fourth quadrant;
θ = 360-30
θ = 330°
Since 180° = πrad
330° = 330π/180
330° = 11π/6 rad
Hence the solutions are 5π/6 rad and 11π/6 rad.
Answer:
The Taylor series of f(x) around the point a, can be written as:

Here we have:
f(x) = 4*cos(x)
a = 7*pi
then, let's calculate each part:
f(a) = 4*cos(7*pi) = -4
df/dx = -4*sin(x)
(df/dx)(a) = -4*sin(7*pi) = 0
(d^2f)/(dx^2) = -4*cos(x)
(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4
Here we already can see two things:
the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.
so we only will work with the even powers of the series:
f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....
So we can write it as:
f(x) = ∑fₙ
Such that the n-th term can written as:

Answer:
y = 1/4 (3 s - 8)
Step-by-step explanation:
hopes this helps:)