Find the surface area of the solid generated by revolving a lemniscate 360°
r^2=a^2 cos2theta

20

step-by-step explanation:

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.

the formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. if a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)