Could someone please help me with this answer please thank you

1 answer:

maria [59]3 years ago

6 0

The bottom answer, as the x coordinate of c to c’ is five across

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ANTONII [103]

9514 1404 393

Answer:

(a) 25°

Step-by-step explanation:

The angle marked 93° is the average of the arcs intercepted by the secant lines.

(161° +AD)/2 = 93°

161° +AD = 186° . . . . . multiply by 2

AD = 25° . . . . . . . . . . .subtract 161°

4 0

3 years ago

ankoles [38]

Answer:

D. undefined

General Formulas and Concepts:

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Trig Derivative: $\displaystyle \frac{d}{dx}[sinu] = u'cosu$

Derivatives of Parametrics: $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Step-by-step explanation:

Step 1: Define

$\displaystyle \frac{dx}{dt} = 5$

$\displaystyle \frac{dy}{dt} = sin(t^2)$

Step 2: Differentiate

[x Derivative] Basic Power Rule: $\displaystyle \frac{d^2x}{dt^2} = 0$
[y Derivative] Trig Derivative [Chain Rule]: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot \frac{d}{dt}[t^2]$
[y Derivative] Basic Power Rule: $\displaystyle \frac{d^2y}{dt^2} = cos(t^2) \cdot 2t^{2 - 1}$
[y Derivative] Simplify: $\displaystyle \frac{d^2y}{dt^2} = 2tcos(t^2)$
[Derivative] Rewrite: $\displaystyle \frac{d^2y}{dx^2} = \frac{2tcos(t^2)}{0}$