bearing in mind that perpendicular lines have negative reciprocal slopes, so
so we're really looking for a line whose slope is 3 and runs through (1,5)
<span>−2/3(12c−9)+14c
= -8c + 6 + 14c
= 6c + 6</span>
Answer:
f(x) = 2π (11 + sin x) √(1 + cos²x)
Step-by-step explanation:
Surface area of a curve rotated about y = a is:
S = ∫ 2π (y − a) ds,
where ds = √(1 + (dy/dx)²) dx.
y = 6 + sin x, and a = -5. dy/dx = cos x. Plugging in:
S = ∫₀²ᵖⁱ 2π (6 + sin x − -5) √(1 + cos²x) dx
S = ∫₀²ᵖⁱ 2π (11 + sin x) √(1 + cos²x) dx
Therefore, f(x) = 2π (11 + sin x) √(1 + cos²x).
Given:
The graph of a trend line.
To find:
The equation of the trend line.
Solution:
From the given graph it is clear that the trend line passes through the points (0,4) and (9,7). So, the equation of the trend line is:
Therefore, the equation of the trend line is .
6,000,000,000+4,000,000+700,000+20,000+80=6,004,720,080