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3 years ago

A cube and a square pyramid were joined to form the composite solid.

Mathematics

1 answer:

Yakvenalex [24]3 years ago

5 0

We know that
total surface area of the composite solid=surface area of the pyramid+surface area of the cube

surface area of the pyramid=4*area of lateral triangle side
area of lateral triangle side=12*9/2----> 54 in²
surface area of the pyramid=4*54---> 216 in²

surface area of the cube=5*[12*12]----> 720 in²

total surface area of the composite solid=216 in²+720 in²---> 936 in²

the answer is the option
936 square inches

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Daniel [21]

Answer:

$\displaystyle \frac{1}{144}arctan(\frac{9x}{2}) + \frac{x}{8(81x^2 + 4)} + C$

General Formulas and Concepts:

Alg I

Terms/Coefficients
Factor
Exponential Rule [Dividing]: $\displaystyle \frac{b^m}{b^n} = b^{m - n}$

Pre-Calc

[Right Triangle Only] Pythagorean Theorem: a² + b² = c²

a is a leg
b is a leg
c is hypotenuse

Trigonometric Ratio: $\displaystyle sec(\theta) = \frac{1}{cos(\theta)}$

Trigonometric Identity: $\displaystyle tan^2\theta + 1 = sec^2\theta$

TI: $\displaystyle sin(2x) = 2sin(x)cos(x)$

TI: $\displaystyle cos^2(\theta) = \frac{cos(2x) + 1}{2}$

Calc

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

IP [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

U-Trig Substitution: x² + a² → x = atanθ

Step-by-step explanation:

Step 1: Define

$\displaystyle \int {\frac{dx}{(81x^2 + 4)^2}}$

Step 2: Identify Sub Variables Pt.1

Rewrite integral [factor expression]:

$\displaystyle \int {\frac{dx}{[(9x)^2 + 4]^2}}$

Identify u-trig sub:

$\displaystyle x = atan\theta\\9x = 2tan\theta \rightarrow x = \frac{2}{9}tan\theta\\dx = \frac{2}{9}sec^2\theta d\theta$

Later, back-sub θ (integrate w/ respect to x):

$\displaystyle tan\theta = \frac{9x}{2} \rightarrow \theta = arctan(\frac{9x}{2})$

Step 3: Integrate Pt.1

[Int] Sub u-trig variables: $\displaystyle \int {\frac{\frac{2}{9}sec^2\theta}{[(2tan\theta)^2 + 4]^2}} \ d\theta$
[Int] Rewrite [Int Prop - MC]: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[(2tan\theta)^2 + 4]^2}} \ d\theta$
[Int] Evaluate exponents: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[4tan^2\theta + 4]^2}} \ d\theta$
[Int] Factor: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[4(tan^2\theta + 1)]^2}} \ d\theta$
[Int] Rewrite [TI]: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{[4sec^2\theta]^2}} \ d\theta$
[Int] Evaluate exponents: $\displaystyle \frac{2}{9} \int {\frac{sec^2\theta}{16sec^4\theta} \ d\theta$
[Int] Rewrite [Int Prop - MC]: $\displaystyle \frac{1}{72} \int {\frac{sec^2\theta}{sec^4\theta} \ d\theta$
[Int] Divide [ER - D]: $\displaystyle \frac{1}{72} \int {\frac{1}{sec^2\theta} \ d\theta$
[Int] Rewrite [TR]: $\displaystyle \frac{1}{72} \int {cos^2\theta} \ d\theta$
[Int] Rewrite [TI]: $\displaystyle \frac{1}{72} \int {\frac{cos(2\theta) + 1}{2}} \ d\theta$
[Int] Rewrite [Int Prop - MC]: $\displaystyle \frac{1}{144} \int {cos(2\theta) + 1} \ d\theta$
[Int] Rewrite [Int Prop - A/S]: $\displaystyle \frac{1}{144} [\int {cos(2\theta) \ d\theta + \int {1} \ d\theta]$

Step 4: Identify Sub Variables Pt.2

Determine u-sub for trig int:

u = 2θ

du = 2dθ

Step 5: Integrate Pt.2

[Ints] Rewrite [Int Prop - MC]: $\displaystyle \frac{1}{144} [\frac{1}{2} \int {2cos(2\theta) \ d\theta + \int {1 \theta ^0} \ d\theta]$
[Int] U-Sub: $\displaystyle \frac{1}{144} [\frac{1}{2} \int {cos(u) \ du + \int {1 \theta ^0} \ d\theta]$
[Ints] Integrate [Trig/Int Rule - RPR]: $\displaystyle \frac{1}{144} [\frac{1}{2} sin(u) + \theta + C]$
[Expression] Back Sub: $\displaystyle \frac{1}{144} [\frac{1}{2} sin(2 \theta) + arctan(\frac{9x}{2}) + C]$
[Exp] Rewrite [TI]: $\displaystyle \frac{1}{144} [\frac{1}{2}(2sin(\theta)cos(\theta)) + arctan(\frac{9x}{2}) + C]$
[Exp] Multiply: $\displaystyle \frac{1}{144} [sin(\theta)cos(\theta) + arctan(\frac{9x}{2}) + C]$
[Exp] Back Sub: $\displaystyle \frac{1}{144} [sin(arctan(\frac{9x}{2}))cos(arctan(\frac{9x}{2})) + arctan(\frac{9x}{2}) + C]$

Step 6: Triangle

Find trig values:

$\displaystyle tan\theta = \frac{9x}{2}$

$\displaystyle \theta = arctan(\frac{9x}{2})$

tanθ = opposite / adjacent; solve hypotenuse of right triangle, determine trig ratios:

sinθ = opposite / hypotenuse

cosθ = adjacent / hypotenuse

Leg a = 2

Leg b = 9x

Leg c = ?

Sub variables [PT]: $\displaystyle 2^2 + (9x)^2 = c^2$
Evaluate exponents: $\displaystyle 4 + 81x^2 = c^2$
[Equality Property] Square root both sides: $\displaystyle \sqrt{4 + 81x^2} = c$
Rewrite: $c = \sqrt{81x^2 + 4}$

Substitute into trig ratios:

$\displaystyle sin\theta = \frac{9x}{\sqrt{81x^2 + 4}}$

$\displaystyle cos\theta = \frac{2}{\sqrt{81x^2 + 4}}$

Step 7: Integrate Pt.3

[Exp] Sub variables [TR]: $\displaystyle \frac{1}{144} [\frac{9x}{\sqrt{81x^2 + 4}} \cdot \frac{2}{\sqrt{81x^2 + 4}} + arctan(\frac{9x}{2}) + C]$
[Exp] Multiply: $\displaystyle \frac{1}{144} [\frac{18x}{81x^2 + 4} + arctan(\frac{9x}{2}) + C]$
[Exp] Distribute: $\displaystyle \frac{1}{144}arctan(\frac{9x}{2}) + \frac{x}{8(81x^2 + 4)} + C$

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2 years ago