Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 4 + ln (t), y = t^2 + 6, (4, 7)

y = 2x − 1

Step-by-step explanation:

By eliminating the parameter, first solve for t:

x = 4 + ln(t)

x − 4 = ln(t)

e^(x − 4) = t

Substitute:

y = t² + 6

y = (e^(x − 4))² + 6

y = e^(2x − 8) + 6

Taking derivative using chain rule:

dy/dx = e^(2x − 8) (2)

dy/dx = 2 e^(2x − 8)

Evaluating at x = 4:

dy/dx = 2 e^(8 − 8)

dy/dx = 2

Writing equation of line using point-slope form:

y − 7 = 2 (x − 4)

y = 2x − 1

Now, without eliminating the parameter, take derivative with respect to t:

x = 4 + ln(t)

dx/dt = 1/t

y = t² + 6

dy/dt = 2t

Finding dy/dx:

dy/dx = (dy/dt) / (dx/dt)

dy/dx = (2t) / (1/t)

dy/dx = 2t²

At the point (4, 7), t = 1.  Evaluating the derivative:

dy/dx = 2(1)²

dy/dx = 2

Writing equation of line using point-slope form:

y − 7 = 2 (x − 4)

y = 2x − 1

1,645,493,059.35

step-by-step explanation:

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq