78+24-11+56+23-117+1
1 answer:
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First, rewrite the equation in standard form.
The center-radius form of the circle equation<span> is in the format:
(x – h)^</span>2<span> + (y – k)^</span>2<span> = r^</span>2
<span>with the center being at the </span>point<span> (h, k) and the radius being "r".
</span>
(x-3)^2 + (y+4)^2 = 81
From here, you can determine the center and radius. The center is at (3,-4) and the radius is 9.
The center-radius form of the circle equation<span> is in the format:
(x – h)^</span>2<span> + (y – k)^</span>2<span> = r^</span>2
<span>with the center being at the </span>point<span> (h, k) and the radius being "r".
</span>
(x-3)^2 + (y+4)^2 = 81
From here, you can determine the center and radius. The center is at (3,-4) and the radius is 9.
Answer:
The graph is sketched by considering the integral. The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
Step-by-step explanation:
We sketch the integral ∫π/40∫6/cos(θ)0f(r,θ)rdrdθ. We consider the inner integral which ranges from r = 0 to r = 6/cosθ. r = 0 is located at the origin and r = 6/cosθ is located on the line x = 6 (since x = rcosθ here x= 6)extends radially outward from the origin. The outer integral ranges from θ = 0 to θ = π/4. This is a line from the origin that intersects the line x = 6 ( r = 6/cosθ) at y = 1 when θ = π/2 . The graph is the region bounded by the origin, the line x = 6, the line y = x/6 and the x-axis.
