<span>The name of the shape graphed by the function r ^ 2 = 9
cos (2 theta) is called the “<u>lemniscate</u>”. A lemniscate is a
plane curve with a feature shape which consists of two loops that meet at a
central point. The curve is also sometimes called as the lemniscate of
Bernoulli. </span>
Explanation:
The
period of coskθ is 2π/k. In this case, k = 2 therefore the
period is π.
r ^ 2 = 9 cos 2θ ≥0 → cos 2θ ≥0. So easily
one period can be chosen as θ ∈
[0, π] wherein cos 2θ ≥0.
As cos(2(−θ)) = cos2θ, the graph is symmetrical about the initial line.
Also,
as cos (2(pi-theta) = cos 2theta, the graph is symmetrical about the
vertical θ = π/2
A
Table for half period [0,π4/] is
adequate for the shape in Quarter1
Use symmetry for the other three quarters:
(r, θ) : (0,3)(3/√√2,π/8)(3√2/2,π/6)(0,π/4<span>)</span>
Convert what is inside the parenthesis to sixths.
(3/6+2/6-1/6)^3+2/3x4/5
(4/6)^3+2/3x4/5
(2/3)^3+12/15
8/27+12/15
(8x12/27x15)
96/405
32/135
Don't know if that's right though
-2
I just took this class for Algebra 1... it would be -2
y=mx+b
this is the equation you will use for these problems. the "b" in the equation represents the y intercept. In this specific problem -2 is the y intercept
2 is not correct
<span>Use the function below to find f(–4). f(x) = 2x</span>
Answer:
Step-by-step explanation:
3(n+4)=2n-8
3n+12=2n-8
5n= -20
n= -4
