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Maurinko [17]
2 years ago
15

What is a formula for the nth term of the given sequence? 19,23,27...

Mathematics
1 answer:
marin [14]2 years ago
7 0
+4 next terms are 33,37,41,44,47,50 and so on.....
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Step-by-step explanation: 5+4+3+2 + 1 = 15

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A: x/4 +1 = -3 B: x+4 = -12 1) How can we get Equation B from Equation A?
Sveta_85 [38]

Answer:In the equation y = mx + b for a straight line, the number m is called the slope of the line. Let x = 0, then y = m • 0 + b, so y = b. The number b is the coordinate on the y-axis where the graph crosses the y-axi

Step-by-step explanation:

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3 years ago
What is 6(m + 4) =_m +_
Zielflug [23.3K]

Answer:

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Step-by-step explanation:

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Find the derivative of
kirill [66]

Answer:

\displaystyle y'(1, \frac{3}{2}) = -3

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<u> </u>

<u>Algebra I</u>

  • Coordinates (x, y)
  • Functions
  • Function Notation
  • Terms/Coefficients
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

Derivatives

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle y = \frac{3}{2x^2}<u />

\displaystyle \text{Point} \ (1, \frac{3}{2})

<u>Step 2: Differentiate</u>

  1. [Function] Rewrite [Exponential Rule - Rewrite]:                                            \displaystyle y = \frac{3}{2}x^{-2}
  2. Basic Power Rule:                                                                                             \displaystyle y' = -2 \cdot \frac{3}{2}x^{-2 - 1}
  3. Simplify:                                                                                                             \displaystyle y' = -3x^{-3}
  4. Rewrite [Exponential Rule - Rewrite]:                                                              \displaystyle y' = \frac{-3}{x^3}

<u>Step 3: Solve</u>

  1. Substitute in coordinate [Derivative]:                                                              \displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1^3}
  2. Evaluate exponents:                                                                                         \displaystyle y'(1, \frac{3}{2}) = \frac{-3}{1}
  3. Divide:                                                                                                               \displaystyle y'(1, \frac{3}{2}) = -3

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

Unit: Derivatives

Book: College Calculus 10e

6 0
2 years ago
Evaluate the algebraic expression.<br> y^2+x+y when x=-3 and y=-4
Amanda [17]

Step-by-step explanation:

you put the values in place of the variable names and calculate.

y² + x + y

(-4)² + -3 + -4 = 16 - 3 - 4 = 16 - 7 = 9

5 0
2 years ago
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