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KengaRu [80]
3 years ago
9

Freee pointsss iggg :)

Mathematics
2 answers:
OlgaM077 [116]3 years ago
8 0
Yayyy thanks a lottt can i also get brainliest
andriy [413]3 years ago
8 0

Answer:

thank you

Step-by-step explanation:

8x8=88 im so smart

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Given BD and AC is a diameter of the circle:

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1. measure of the arc BA

So,

the measure of the minor arc BA = 44

The measure of the major arc BA = 360 - 44 = 316

2. the measure of the arc ACB = 360 - 44 = 316

3. The arc BCD is a semicircle of the circle

So, the measure of the arc BCD = 180

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10 months ago
At any point in time, there could be bicycles, tricycles, and
Aleksandr-060686 [28]

Answer:

There may be 1 or 3 tricycles in the parking lot.

Step-by-step explanation:

Since at any point in time, there could be bicycles, tricycles, and cars in the school parking lot, and today, there are 53 wheels in total, if there are 15 bicycles, tricycles, and cars in total, to determine how many tricycles could be in the parking lot, the following calculation must be performed:

13 x 4 + 1 x 3 + 1 x 2 = 57

11 x 4 + 1 x 3 + 3 x 2 = 53

10 x 4 + 3 x 3 + 2 x 2 = 53

8 x 4 + 5 x 3 + 2 x 2 = 51

10 x 2 + 1 x 3 + 4 x 4 = 39

9 x 3 + 1 x 2 + 5 x 4 = 49

Therefore, there may be 1 or 3 tricycles in the parking lot.

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Write the formula for the area of a triangle, A=1/2bh in terms of h. Find the height of a triangle when A = 18 in. squared and b
Elena L [17]

The height of the given triangle is  

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Distance between (-4,0) and (2,2)
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(4,2) is is the distance from these coordinates
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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
sattari [20]

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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