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Kobotan [32]
3 years ago
10

Find two vectors v1 and v2 whose sum is 〈−3,4〉〈−3,4〉, where v1 is parallel to 〈3,−3〉〈3,−3〉 while v2 is perpendicular to 〈3,−3〉〈3

,−3〉.
Mathematics
1 answer:
Sati [7]3 years ago
6 0

Answer:

∵ Vector v_1 is parallel to 〈3, -3〉,

⇒ v_1 = x〈3, -3〉 = 〈3x, -3x〉,

Where, x is any scalar,

According to the question,

v_1+v_2 = 〈-3,4〉

⇒ 〈3x, -3x〉+ v_2 = 〈-3,4〉

⇒ v_2 = 〈-3,4〉 - 〈3x,-3x〉

Now, v_2 is perpendicular to 〈3, -3〉,

⇒ v_2.〈3, -3〉 = 0

⇒ ( 〈3,4〉 - 〈3x, -3x〉 ). 〈3, -3〉 = 0

⇒ 〈-3-3x,4+3x〉 . 〈3, -3〉 = 0

⇒ (-3-3x)(3) + (4+3x)(-3) = 0

⇒ - 9 - 9x - 12 - 9x = 0

⇒ -21 - 18x = 0

⇒ x = -\frac{7}{6}

Hence, required vectors are,

v_1 = 〈-7/2, 7/2〉

v_2 =〈3,4〉 - 〈-7/2, 7/2〉 = 〈13/2, 1/2〉

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Jack looks at a clock tower from a distance and determines that the angle of elevation of the top of the tower is 40°. John, who
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tan(40) = (y + 1.5) / (x + 20)

y + 1.5 = (x + 20)[tan(40)]

y = (x + 20)[tan(40)] - 1.5

tan(60) = (y + 1.5) / x

y + 1.5 = (x)[tan(60)]

y = (x)[tan(60)] - 1.5

(x)[tan(60)] - 1.5 = (x + 20)[tan(40)] - 1.5

(x)[tan(60)] = (x + 20)[tan(40)]

(x)[tan(60)] = (x)[tan(40)] + (20)[tan(40)]

(x)[tan(60)] - (x)[tan(40)] = (20)[tan(40)]

(x)[tan(60) - tan(40)] = (20)[tan(40)]

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Without multiplying, determine the sign of the product (−247,634)(183,758).
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The profit P (in thousands of dollars) for a company spending an amount s (in thousands of dollars on advertising is
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Answer:

The company should spend $40 to yield a maximum profit.

The point of diminishing returns is (40, 3600).

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  

Equality Properties

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

<u>Algebra I</u>

Coordinate Planes

  • Coordinates (x, y) → (s, P)

Functions

  • Function Notation

Terms/Coefficients

  • Factoring/Expanding

Quadratics

<u>Algebra II</u>

Coordinate Planes

  • Maximums/Minimums

<u>Calculus</u>

Derivatives

  • Derivative Notation

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

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

1st Derivative Test - tells us where on the function f(x) does it have a relative maximum or minimum

  • Critical Numbers

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle P = \frac{-1}{10}s^3 + 6s^2 + 400

<u>Step 2: Differentiate</u>

  1. [Function] Derivative Property [Addition/Subtraction]:                               \displaystyle P' = \frac{dP}{ds} \bigg[ \frac{-1}{10}s^3 \bigg] + \frac{dP}{ds} [ 6s^2 ] + \frac{dP}{ds} [ 400 ]
  2. [Derivative] Rewrite [Derivative Property - Multiplied Constant]:               \displaystyle P' = \frac{-1}{10} \frac{dP}{ds} \bigg[ s^3 \bigg] + 6 \frac{dP}{ds} [ s^2 ] + \frac{dP}{ds} [ 400 ]
  3. [Derivative] Basic Power Rule:                                                                     \displaystyle P' = \frac{-1}{10}(3s^2) + 6(2s)
  4. [Derivative] Simplify:                                                                                     \displaystyle P' = -\frac{3s^2}{10}  + 12s

<u>Step 3: 1st Derivative Test</u>

  1. [Derivative] Set up:                                                                                       \displaystyle 0 = -\frac{3s^2}{10}  + 12s
  2. [Derivative] Factor:                                                                                       \displaystyle 0 = \frac{-3s(s - 40)}{10}
  3. [Multiplication Property of Equality] Isolate <em>s </em>terms:                                   \displaystyle 0 = -3s(s - 40)
  4. [Solve] Find quadratic roots:                                                                         \displaystyle s = 0, 40

∴ <em>s</em> = 0, 40 are our critical numbers.

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

  1. [Function] Substitute in <em>s</em> = 0:                                                                       \displaystyle P(0) = \frac{-1}{10}(0)^3 + 6(0)^2 + 400
  2. [Order of Operations] Evaluate:                                                                   \displaystyle P(0) = 400
  3. [Function] Substitute in <em>s</em> = 40:                                                                     \displaystyle P(40) = \frac{-1}{10}(40)^3 + 6(40)^2 + 400
  4. [Order of Operations] Evaluate:                                                                   \displaystyle P(40) = 3600

We see that we will have a bigger profit when we spend <em>s</em> = $40.

∴ The maximum profit is $3600.

∴ The point of diminishing returns is ($40, $3600).

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

Unit: Differentiation (Applications)

5 0
2 years ago
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