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UNO [17]
3 years ago
7

If someone wanted to become a chef how would they achieve their goals? Need very long response plz!

Mathematics
1 answer:
Licemer1 [7]3 years ago
7 0

Answer:

if they wanted to achive their goals they could maybe go to a resturaun t and ytalk to a chef for some tips. Then the person could practice at home or wherever else and imporve their skills. Finally if they make mistakes you caould ask somebody you know to help you.

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Factor the GCF: −12x3y5 − 9x2y2 + 12xy3
Luba_88 [7]
<span>−12x3y5 − 9x2y2 + 12xy3</span>
-3xy^2(4x^2y^3+3x-4y)
6 0
4 years ago
The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.
puteri [66]

Answer:

\displaystyle \frac{dA}{dt} = 102 \ m^2/day

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<u> </u>

<u>Geometry</u>

Area of a Rectangle: A = lw

  • l is length
  • w is width

<u>Calculus</u>

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

Derivative Rule [Product Rule]:                                                                              \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle l = 10 \ meters<u />

<u />\displaystyle \frac{dl}{dt} = 4 \ m/day<u />

<u />\displaystyle w = 23 \ meters<u />

<u />\displaystyle \frac{dw}{dt} = 1 \ m/day<u />

<u />

<u>Step 2: Differentiate</u>

  1. [Area of Rectangle] Product Rule:                                                                 \displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}

<u>Step 3: Solve</u>

  1. [Rate] Substitute in variables [Derivative]:                                                    \displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)
  2. [Rate] Multiply:                                                                                                \displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day
  3. [Rate] Add:                                                                                                      \displaystyle \frac{dA}{dt} = 102 \ m^2/day

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

Unit: Implicit Differentiation

Book: College Calculus 10e

8 0
3 years ago
Use a linear approximation (or differentials) to estimate the given number. (Round your answer to five decimal places.) 3 217
Soloha48 [4]

Answer:

f(216) \approx 6.0093

Step-by-step explanation:

Given

\sqrt[3]{217}

Required

Solve

Linear approximated as:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

Take:

x = 216; \triangle x= 1

So:

f(x) = \sqrt[3]{x}

Substitute 216 for x

f(x) = \sqrt[3]{216}

f(x) = 6

So, we have:

f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

f(215 + 1) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot f'(x)

To calculate f'(x);

We have:

f(x) = \sqrt[3]{x}

Rewrite as:

f(x) = x^\frac{1}{3}

Differentiate

f'(x) = \frac{1}{3}x^{\frac{1}{3} - 1}

Split

f'(x) = \frac{1}{3} \cdot \frac{x^\frac{1}{3}}{x}

f'(x) = \frac{x^\frac{1}{3}}{3x}

Substitute 216 for x

f'(216) = \frac{216^\frac{1}{3}}{3*216}

f'(216) = \frac{6}{648}

f'(216) = \frac{3}{324}

So:

f(216) \approx 6  +1 \cdot f'(x)

f(216) \approx 6  +1 \cdot \frac{3}{324}

f(216) \approx 6  + \frac{3}{324}

f(216) \approx 6  + 0.0093

f(216) \approx 6.0093

6 0
3 years ago
Solve the system of equations below.
Ostrovityanka [42]
C - no solution
you take the second solution and times the entire equation by -1
then simply
and you get 0x + 0y = 20
5 0
4 years ago
Read 2 more answers
What is the equation to this?
Fiesta28 [93]
THE EQUATION TO THIS PROBLEM IS y=2x+12
7 0
3 years ago
Read 2 more answers
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