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.
#SPJ2
5x + 3(x – 5) = 6x + 2x – 15
Part A: Solve the equation and write the number of solutions. Show all the steps. (6 points)
Part B: Name one property you used to solve this equation. (4 points)
Hi there
5x+3(x-5)=6x+2x-15
Simplify both sides of the equation
5x+3(x-5)=6x+2x-15
Distribute
5x+3x-15=6x+2x-15
Combine like terms
(5x+3x)-15= (6x+2x)-15
8x-15=8x-15
Subtract 8x from both sides
8x-15-8x=8x-15-8x
-15=-15
Add 15 to both sides
-15+15=-15+15
0=0
All real numbers are solutions
To solve this equation I used " The distributive property"
I hope that's help !
Good night !
(5,2)
5=x
y=2
5-3=2
hope that helps