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lidiya [134]
3 years ago
11

The right rectangular prism below is made of equal-sized cubes the side length of each cube is 2 1/2 inches . What’s is the volu

me in cubic images of the right rectangular prism?

Mathematics
1 answer:
vovikov84 [41]3 years ago
4 0
We know that

volume of one cube=b³
where 
b is the length side of cube
b=2 1/2 in-----> (2*2+1)/2----> 5/2 in
volume of one cube=(5/2)³-----> 125/8 in³

volume of the prism=[volume of one cube]*5*2*4
volume of the prism=[volume of one cube]*40
volume of the prism=[125/8]*40----> 625 in³

the answer is
the volume of the prism is equal to the volume of 40 cubes
the volume of the prism is equal to 625 in³

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What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)
zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

  • Integrals
  • Definite Integrals
  • Integration Constant C

Arc Length Formula [Parametric]:                                                                         \displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.

Interval [0, π]

<u>Step 2: Find Arc Length</u>

  1. [Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]:         \displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.
  2. Substitute in variables [Arc Length Formula - Parametric]:                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx
  3. [Integrand] Simplify:                                                                                       \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx
  4. [Integral] Evaluate:                                                                                         \displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0
3 years ago
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