Answer:
1: 3 miles
2: 1 mile
Step-by-step explanation:
2
sin
x
at the point (π/6,1)
π
6
1
.
The equation of this tangent line can be written in the form y=mx+b
y
m
x
b
where
To find the equation of the tangent line to the curve y=2sinx at the point (π/6,1), we take the derivative to find the slope and then use the point-slope form of the line equation. The result is y = √3x + 1 - √3π/6.
The subject of this question is calculus and focuses specifically on finding the equation of the tangent line to the curve y=2sinx at a given point. To do this, we use the formula y=mx+b.
Firstly, the slope of the tangent line is obtained by taking the derivative of the function at the point of tangency. The derivative of y=2sinx is y'=2cosx. For the given point (π/6,1), the slope (m) would be 2cos(π/6) = √3.
Secondly, we use the point-slope form of the line equation to find b. Inserting the values of the slope (m) and the given point into the equation, we get 1 = √3(π/6) + b. Solving for b gives b = 1 - √3π/6.
Finally, the equation of the tangent line is y = √3x + 1 - √3π/6.
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[42 ÷ (2 + 3 • 2)] + [4 • (36 – 52)]
I really need help!!!
B. The sides of the polygon are chords of the circle.
C. The vertices of the polygon are on the circumference of the circle.
D. The sides of the polygon are tangents of the circle.