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gayaneshka [121]
3 years ago
5

Olivia goes to Egypt and saw a model of triangle shaped pyramid. She enlarging using a scale factor of 1.2. What will be the per

imeter and area of the new triangle pyramid?
Mathematics
1 answer:
ivann1987 [24]3 years ago
8 0
If you are talking about the area and perimeter of one of the triangular faces...
8 x 1.2 = 9.6
6 x 1.2 = 7.2

Perimeter = 9.6 + 7.2 + 7.2 = 24
Area = Linear Scale Factor²
1.2² = 1.44
7 x 1.44 = 10.08 (units)²
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True or false: arc measure is the same length as arc length
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A marble jar contains 1 blue, 2 green, 3 red, and 5 yellow marbles. If a single marble is drawn at random from the jar, what is
AleksAgata [21]

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4/11

Step-by-step explanation:

1 blue, 2 green, 3 red, and 5 yellow marbles.= 11 marbles

P( blue or red) = number of blue or red / total

                        = 4/11

5 0
3 years ago
The federal government would like to test the hypothesis that the average age of men filing for Social Security is higher than t
mixas84 [53]

Answer:

The 90% confidence interval is  -0.3433<  \mu_1 - \mu_2 < 2.1433

Step-by-step explanation:

From the question we are told that

   The sample mean for men is  \= x_1 = 64.5 \ years

    The sample mean for women is  \= x_2 = 63.6 \ years

     The sample size for men  is   n_1 = 35

      The sample size for women is  n_2 = 39

      The standard deviation for men is s_1 = 3.0

       The standard deviation for women is  s_2 = 3.5

From the question we are told the confidence level is  90% , hence the level of significance is    

      \alpha = (100 - 90 ) \%

=>   \alpha = 0.10

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.645

Generally the standard error is mathematically represented as

     SE = \sqrt{\frac{s_1^2 }{n_1}  + \frac{s_2^2}{n_2} }

=>  SE = \sqrt{\frac{3^2 }{35}  + \frac{3.5^2}{39} }

=>  SE = 0.7558

Generally the margin of error is mathematically represented as

       E =  Z_{\frac{\alpha }{2} } *  SE

=>     E =  1.645  *  0.7558

=>     E =  1.2433

Generally 90% confidence interval is mathematically represented as  

      (\= x_1 - \= x_2) -E <  \mu_1 - \mu_2 <  (\= x_1 - \= x_2) -E

=>    (64.5 - 63.6) -1.2433<  \mu_1 - \mu_2

=>    -0.3433<  \mu_1 - \mu_2 < 2.1433

7 0
2 years ago
The third-degree Taylor polynomial about x = 0 of In(1 - x) is
gizmo_the_mogwai [7]

Answer:

\displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{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>

  • Functions
  • Function Notation

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Rule [Quotient Rule]:                                                                                \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

MacLaurin/Taylor Polynomials

  • Approximating Transcendental and Elementary functions
  • MacLaurin Polynomial:                                                                                     \displaystyle P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + ... + \frac{f^{(n)}(0)}{n!}x^n
  • Taylor Polynomial:                                                                                            \displaystyle P_n(x) = \frac{f(c)}{0!} + \frac{f'(c)}{1!}(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \frac{f'''(c)}{3!}(x - c)^3 + ... + \frac{f^{(n)}(c)}{n!}(x - c)^n

Step-by-step explanation:

*Note: I will not be showing the work for derivatives as it is relatively straightforward. If you request for me to show that portion, please leave a comment so I can add it. I will also not show work for elementary calculations.

<u />

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = ln(1 - x)

Center: x = 0

<em>n</em> = 3

<u>Step 2: Differentiate</u>

  1. [Function] 1st Derivative:                                                                                  \displaystyle f'(x) = \frac{1}{x - 1}
  2. [Function] 2nd Derivative:                                                                                \displaystyle f''(x) = \frac{-1}{(x - 1)^2}
  3. [Function] 3rd Derivative:                                                                                 \displaystyle f'''(x) = \frac{2}{(x - 1)^3}

<u>Step 3: Evaluate Functions</u>

  1. Substitute in center <em>x</em> [Function]:                                                                     \displaystyle f(0) = ln(1 - 0)
  2. Simplify:                                                                                                             \displaystyle f(0) = 0
  3. Substitute in center <em>x</em> [1st Derivative]:                                                             \displaystyle f'(0) = \frac{1}{0 - 1}
  4. Simplify:                                                                                                             \displaystyle f'(0) = -1
  5. Substitute in center <em>x</em> [2nd Derivative]:                                                           \displaystyle f''(0) = \frac{-1}{(0 - 1)^2}
  6. Simplify:                                                                                                             \displaystyle f''(0) = -1
  7. Substitute in center <em>x</em> [3rd Derivative]:                                                            \displaystyle f'''(0) = \frac{2}{(0 - 1)^3}
  8. Simplify:                                                                                                             \displaystyle f'''(0) = -2

<u>Step 4: Write Taylor Polynomial</u>

  1. Substitute in derivative function values [MacLaurin Polynomial]:                 \displaystyle P_3(x) = \frac{0}{0!} + \frac{-1}{1!}x + \frac{-1}{2!}x^2 + \frac{-2}{3!}x^3
  2. Simplify:                                                                                                             \displaystyle P_3(x) = -x - \frac{x^2}{2} - \frac{x^3}{3}

Topic: AP Calculus BC (Calculus I/II)

Unit: Taylor Polynomials and Approximations

Book: College Calculus 10e

5 0
3 years ago
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