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igomit [66]
2 years ago
6

The mass of a basket containing 3 balls was 3.64 kilograms. The mass of the same basket and 5 balls is 5.26 kilograms. Find the

mass of the empty basket in kilograms.
a. the mass of a empty basket is ___
b. explain your answer (in short)

please help
Mathematics
2 answers:
aksik [14]2 years ago
7 0

Answer:

mass of empty basket: 1.21 kg

Explanation:

the mass of basket is constant. Let the mass of basket be "b" and mass of each balls be "m"

Make equations:

  • b + 3(m) = 3.64  ___ equation 1
  • b + 5(m) = 5.26  ___ equation 2

<u>solve them simultaneously,</u>

  • 3.64 - 3(m) = 5.26 - 5(m)
  • -3(m) + 5(m) = 5.26 - 3.64
  • 2(m) = 1.62
  • m = 0.81 kg

<u>mass of basket,</u>

  • b = 3.64 - 3(m)
  • 3.64 - 3(0.81 )
  • 1.21 kg
jenyasd209 [6]2 years ago
4 0
  1. Bucket be x
  2. balls be y

Now

  • x+3y=3.64=>x=3.64-3y--(1)
  • x+5y=5.26=>x=5.26-5y

So

  • 3.64-3y=5.26-5y
  • -3y+5y=5.26-3.64
  • 2y=1.62
  • y=0.81kg

So

  • x=3.64-3(0.81)=3.64-2.43=1.21kg
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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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