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Wittaler [7]
2 years ago
15

HELPP!!! 100 POINTS AND BRAINLIEST!! Sketch the graph of the function y = 20x − x2, and approximate the area under the curve in

the interval [0, 20] by dividing the area into the given numbers of rectangles.
Part A
Use five rectangles to approximate the area under the curve.
Part B
Use 10 rectangles to approximate the area under the curve.
Part C
Calculate the area under the curve using rectangles as their number becomes arbitrarily large (tends to infinity). You do not need to sketch the rectangles.
Mathematics
1 answer:
Arte-miy333 [17]2 years ago
5 0

Answer:

Hello,

Step-by-step explanation:

Part A: see picture

Part B: see picture

Part C:

You must remenber :

\displaystyle \sum_{i=1}^{n} i= \dfrac{n*(n+1)}{2} \\\\\displaystyle \sum_{i=1}^{n} i^2= \dfrac{n*(n+1)(2n+1)}{6} \\\\\\\int\limits^{20}_0 {(20\ x-x^2)} \, dx =[\dfrac{20\ x^2}{2} -\dfrac{x^3}{3} ]^{20}_0=\dfrac{20^3}{3} =1333.\overline{3}\\\\\\

We divide the intervalle in n equal parts:

f(x)=20\ x- x^2\\\Delta x=\dfrac{20-0}{n} =\dfrac{20}{n} \\Area\ of\ a\ rectangle\ R_i=\Delta x*f(x_i)\\x_i=0+\Delta x*i=\Delta x*i\\\\\\\displaystyle \int\limits^{20}_0 {(20 \ x-x^2)} \, dx = \lim_{n \to \infty}  \sum_{i=1}^n(\Delta x* f(x_i) )\\\\=\lim_{n \to \infty}  \sum_{i=1}^n(\Delta x* (20*\Delta x*i-(\Delta x*i)^2)\\\\=\lim_{n \to \infty} (20*(\dfrac{20}{n})^2\sum_{i=1}^n(i)-(\dfrac{20}{n})^3\sum_{i=1}^n(i^2)\\\\

\displaystyle =\lim_{n \to \infty}(20*(\dfrac{20}{n})^2*\dfrac{n*(n+1)}{2} -(\dfrac{20}{n})^3*\dfrac{n*(n+1)(2n+1)}{6} \\\\=10*20^2-\dfrac{20^3}{6} *2\\\\\\=4000-\frac{8000}{3} \\\\=\dfrac{4000}{3} \\\\=1333,\overline{3}\\\\

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Answer:

The number that belongs <em>in</em> the green box is equal to 909.

General Formulas and Concepts:
<u>Algebra I</u>

Equality Properties

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

<u>Trigonometry</u>

[<em>Right Triangles Only</em>] Pythagorean Theorem:
\displaystyle a^2 + b^2 = c^2

  • a is a leg
  • b is another leg
  • c is the hypotenuse

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify given variables</em>.

<em>a</em> = 30

<em>b</em> = 3

<em>c</em> = <em>x</em>

<em />

<u>Step 2: Find </u><u><em>x</em></u>

Let's solve for the <em>general</em> equation that allows us to find the hypotenuse:

  1. [Pythagorean Theorem] Square root both sides [Equality Property]:
    \displaystyle \begin{aligned}a^2 + b^2 = c^2 \rightarrow c = \sqrt{a^2 + b^2}\end{aligned}

Now that we have the <em>formula</em> to solve for the hypotenuse, let's figure out what <em>x</em> is equal to:

  1. [Equation] <em>Substitute</em> in variables:
    \displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2}\end{aligned}
  2. <em>Evaluate</em>:
    \displaystyle \begin{aligned}c & = \sqrt{a^2 + b^2} \\x & = \sqrt{30^2 + 3^2} \\& = \boxed{ \sqrt{909} } \\\end{aligned}

∴ the hypotenuse length <em>x</em> is equal to √909 and the number <em>under</em> the square root, our answer, is equal to 909.

___

Learn more about Trigonometry: brainly.com/question/27707750

___

Topic: Trigonometry

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