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kogti [31]
2 years ago
9

A beautiful cake was presented at a wedding reception. The cake measured 13.5 inches tall, including the decorative icing on top

that is one inch thick. If each of the six cake layers measures 1.25 inches tall, how thick is the icing between the layers? Select an Answer A 1 fourth inches B 1 half inches C 2 over 3 inches D 3 over 4 inches E 1 inch
Mathematics
1 answer:
Schach [20]2 years ago
5 0

Using proportions, it is found that the thickness of the icing between the layers, in inches, is of:

E 1 inch

<h3>What is a proportion?</h3>

A proportion is a fraction of a total amount.

In this problem, there are six layers, each measuring 1.25 inches tall, hence the height of these layers is:

6 x 1.25 = 7.5 inches

Then the height of the icing between the layers is of:

13.5 - 7.5 = 6 inches

Six layers, hence each icing is of:

6/6 = 1 inch

More can be learned about proportions at brainly.com/question/24372153

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ankoles [38]

Answer:

D. undefined

General Formulas and Concepts:

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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:

<u>Step 1: Define</u>

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

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

<u>Step 2: Differentiate</u>

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

Anything divided by 0 is undefined.

Topic: AP Calculus BC (Calculus I/II)

Unit: Differentiation with Parametrics

Book: College Calculus 10e

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