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Shkiper50 [21]
3 years ago
15

Find mRT. Assume that segments that appear to be tangent are tangent.

Mathematics
1 answer:
IceJOKER [234]3 years ago
7 0

Answer:

m(arc RT) = 148°

Step-by-step explanation:

From the picture attached,

Segment RT is a chord and segment RS is a tangent of a circle O meeting at R.

By the property of tangent chord angle,

"Angle formed by an intersecting tangent and chord measures half of the intercepted minor arc"

m(∠SRT) = \frac{1}{2}(\text{minor arc RT})

74^0=\frac{1}{2}m(RT)

m(\text{arc RT)}=2(74^0)

                 =148^0

Therefore, measure of minor arc RT is 148°.

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The point of diminishing returns is (40, 3600).

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

Coordinate Planes

  • Coordinates (x, y) → (s, P)

Functions

  • Function Notation

Terms/Coefficients

  • Factoring/Expanding

Quadratics

<u>Algebra II</u>

Coordinate Planes

  • Maximums/Minimums

<u>Calculus</u>

Derivatives

  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

1st Derivative Test - tells us where on the function f(x) does it have a relative maximum or minimum

  • Critical Numbers

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle P = \frac{-1}{10}s^3 + 6s^2 + 400

<u>Step 2: Differentiate</u>

  1. [Function] Derivative Property [Addition/Subtraction]:                               \displaystyle P' = \frac{dP}{ds} \bigg[ \frac{-1}{10}s^3 \bigg] + \frac{dP}{ds} [ 6s^2 ] + \frac{dP}{ds} [ 400 ]
  2. [Derivative] Rewrite [Derivative Property - Multiplied Constant]:               \displaystyle P' = \frac{-1}{10} \frac{dP}{ds} \bigg[ s^3 \bigg] + 6 \frac{dP}{ds} [ s^2 ] + \frac{dP}{ds} [ 400 ]
  3. [Derivative] Basic Power Rule:                                                                     \displaystyle P' = \frac{-1}{10}(3s^2) + 6(2s)
  4. [Derivative] Simplify:                                                                                     \displaystyle P' = -\frac{3s^2}{10}  + 12s

<u>Step 3: 1st Derivative Test</u>

  1. [Derivative] Set up:                                                                                       \displaystyle 0 = -\frac{3s^2}{10}  + 12s
  2. [Derivative] Factor:                                                                                       \displaystyle 0 = \frac{-3s(s - 40)}{10}
  3. [Multiplication Property of Equality] Isolate <em>s </em>terms:                                   \displaystyle 0 = -3s(s - 40)
  4. [Solve] Find quadratic roots:                                                                         \displaystyle s = 0, 40

∴ <em>s</em> = 0, 40 are our critical numbers.

<u>Step 4: Find Profit</u>

  1. [Function] Substitute in <em>s</em> = 0:                                                                       \displaystyle P(0) = \frac{-1}{10}(0)^3 + 6(0)^2 + 400
  2. [Order of Operations] Evaluate:                                                                   \displaystyle P(0) = 400
  3. [Function] Substitute in <em>s</em> = 40:                                                                     \displaystyle P(40) = \frac{-1}{10}(40)^3 + 6(40)^2 + 400
  4. [Order of Operations] Evaluate:                                                                   \displaystyle P(40) = 3600

We see that we will have a bigger profit when we spend <em>s</em> = $40.

∴ The maximum profit is $3600.

∴ The point of diminishing returns is ($40, $3600).

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation (Applications)

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