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3 years ago

Find the surface area of the figure shown.

Mathematics

1 answer:

balandron [24]3 years ago

3 0

Answer:

The correct answer is option d. 576 ft^2

Step-by-step explanation:

From the figure we can see that a prism

To find the surface area

Surface area = Base area + area of 2 triangles + Two side area

Base area = length * breadth = 15 * 12 = 180 ft^2

Area of triangle = bh/2 = (12 * 8)/2 = 48 ft^2

Area of two triangle = 48 * 2 = 96 ft^2

Area of one side = 15 * 10 = 150 ft^2

Area of two side = 2 * 150 = 300 ft^2

Surface area = 180 + 96 + 300 = 576 ft^2

The correct answer is option d. 576 ft^2

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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:

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
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:

Step 1: Define

Identify

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

Interval [0, π]

Step 2: Find Arc Length

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