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Kay [80]
3 years ago
12

Find the volume of a pyramid with a square base, where the perimeter of the base is 8.2\text{ cm}8.2 cm and the height of the py

ramid is 13\text{ cm}13 cm. Round your answer to the nearest tenth of a cubic centimeter.
Mathematics
1 answer:
7nadin3 [17]3 years ago
3 0

Answer:

V=18.2cm^3

Step-by-step explanation:

The volume of the pyramid is given by:

V=\frac{A_{b}h}{3}

where A_{b} is the area of the base and h is the height.

We know that the height is:

h=13cm

Thus we need to find the area of the square at the base.

The perimeter of the square is:

p=8.2cm

this means that the length of the sides of the square is :

l=\frac{p}{4}\\ \\l=\frac{8.2cm}{4}\\ \\l=2.05cm

Now we can find the area of the base which the area of the square:

A_{b}=l^2\\A_b=(2.05cm)^2\\A_b=4.2025cm^2

and finally we find the volume:

V=\frac{A_{b}h}{3}

V=\frac{(4.2025cm^2)(13cm)}{3}\\ \\V=\frac{54.6325cm^3}{3}\\ \\V=18.2108cm^3

Which rounded to the nearest tenth of a cubic centimeter is: V=18.2cm^3

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Mariana [72]

Answer:

Full-time worker minimum monthly income = US$ 7,846.40

Part-time worker minimum monthly income  = US$ 7,846.40

Full-time worker hourly wage = US$ 49.04

Part-time worker hourly wage = US$ 98.08

Step-by-step explanation:

1. Let's review the information given to answer the question correctly:

Monthly costs of a 2-adult and 3-children family in the Modesto metro area = US$ 7,846

Full-time worker = 4 weeks of 40-hour per month

Part-time worker = 2 weeks of 40-hour per month

2. Let's calculate the minimum monthly income and hourly wage per worker

Full-time worker hourly wage = Monthly costs/160

Part-time worker hourly wage = Monthly costs/80

Full-time worker hourly wage = 7,846/160

Part-time worker hourly wage = 7,846/80

Full-time worker hourly wage = US$ 49.04

Part-time worker hourly wage = US$ 98.08

Full-time worker minimum monthly income = Hourly wage * 160

Part-time worker minimum monthly income  = Hourly wage * 80

Full-time worker minimum monthly income = US$ 7,846.40

Part-time worker minimum monthly income  = US$ 7,846.40

Step-by-step explanation:

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If anyone knows about definite integrals for calculus then please I request help! I
kicyunya [14]

Answer:

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

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

Basic Power Rule:

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Integration

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Integration Rule [Fundamental Theorem of Calculus 1]:                                     \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Multiplied Constant]:                                                         \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

U-Substitution

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 4x^{-2}
  2. [<em>u</em>] Differentiate [Basic Power Rule, Derivative Properties]:                       \displaystyle du = \frac{-8}{x^3} \ dx
  3. [Bounds] Switch:                                                                                           \displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx
  2. [Integral] U-Substitution:                                                                              \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du
  3. [Integral] Exponential Integration:                                                               \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)
  5. Simplify:                                                                                                         \displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)

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

Unit: Integration

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Step-by-step explanation:

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Answer:

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