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Anna71 [15]
3 years ago
12

Please help with this question.

Mathematics
1 answer:
horsena [70]3 years ago
4 0

Answer:

Apples and Peaches

Step-by-step explanation:

It is convenient to compare the prices for 3 lbs of each of the fruits.

Multiply the price and quantity of Apples by 4:

... 4×(3/4 lb) = 4×$1.59

... 3 lb = $6.36 . . . . for apples

Multiply the price and quantity of Plums by 6:

... 6×(0.5 lb) = 6×$1.15

... 3 lb = $6.90 . . . . for plums

The two lowest-priced fruits are apples (6.36/(3 lb)) and peaches (5.37/(3 lb)).

_____

<em>Comment on</em> the best deal

No criteria are provided for determining the best deal. Though plums are higher-priced, they have somewhat higher polyphenol content than apples or peaches, so may provide the best deal in terms of nutrition per dollar.

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Step-by-step explanation:

As:

f (x) = 3/4 x + 12

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x = 12/1/3/4

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F(x) = e^-x . Find the equation of the tangent to f(x) at x=-1​
natima [27]

Answer:

The <em>equation</em> of the tangent line is given by the following equation:

\displaystyle y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg)

General Formulas and Concepts:

<u>Algebra I</u>

Point-Slope Form: y - y₁ = m(x - x₁)

  • x₁ - x coordinate
  • y₁ - y coordinate
  • m - slope

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

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

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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

Step-by-step explanation:

*Note:

Recall that the definition of the derivative is the <em>slope of the tangent line</em>.

<u>Step 1: Define</u>

<em>Identify given.</em>

<em />\displaystylef(x) = e^{-x} \\x = -1

<u>Step 2: Differentiate</u>

  1. [Function] Apply Exponential Differentiation [Derivative Rule - Chain Rule]:
    \displaystyle f'(x) = e^{-x}(-x)'
  2. [Derivative] Rewrite [Derivative Rule - Multiplied Constant]:
    \displaystyle f'(x) = -e^{-x}(x)'
  3. [Derivative] Apply Derivative Rule [Derivative Rule - Basic Power Rule]:
    \displaystyle f'(x) = -e^{-x}

<u>Step 3: Find Tangent Slope</u>

  1. [Derivative] Substitute in <em>x</em> = 1:
    \displaystyle f'(1) = -e^{-1}
  2. Rewrite:
    \displaystyle f'(1) = \frac{-1}{e}

∴ the slope of the tangent line is equal to  \displaystyle \frac{-1}{e}.

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

  1. [Function] Substitute in <em>x</em> = 1:
    \displaystyle f(1) = e^{-1}
  2. Rewrite:
    \displaystyle f(1) = \frac{1}{e}

∴ our point is equal to  \displaystyle \bigg( 1, \frac{1}{e} \bigg).

Substituting in our variables we found into the point-slope form general equation, we get our final answer of:

\displaystyle \boxed{ y - \frac{1}{e} = \frac{-1}{e} \bigg( x - 1 \bigg) }

∴ we have our final answer.

---

Learn more about derivatives: brainly.com/question/27163229

Learn more about calculus: brainly.com/question/23558817

---

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

Unit: Differentiation

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