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3 years ago

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The value of G for a one-piece 30° x 60° angle iron bracket is 1'/4 inches, and the width of the angle iron is 2 inches. What is

the thickness of the metal, in inches? A. 31/4 inches B. 3/4 inch C. 21/2 inches D. 1/4 inch

1 answer:

Alborosie3 years ago

8 0

Answerb

Step-by-step explanation:

yeal

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2 years ago

Evaluate the interval (Calculus 2)

Darya [45]

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$2 \tan (6x)+2 \sec (6x)+\text{C}$

Step-by-step explanation:

<u>Fundamental Theorem of Calculus</u>

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{12}{1-\sin (6x)}\:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int a\:\text{f}(x)\:\text{d}x=a \int \text{f}(x) \:\text{d}x$\end{minipage}}$

If the terms are multiplied by constants, take them outside the integral:

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)}\:\:\text{d}x$

Multiply by the conjugate of 1 - sin(6x) :

$\implies 12\displaystyle \int \dfrac{1}{1-\sin (6x)} \cdot \dfrac{1+\sin(6x)}{1+\sin(6x)}\:\:\text{d}x$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{1-\sin^2(6x)} \:\:\text{d}x$

$\textsf{Use the identity} \quad \sin^2 x+ \cos^2 x=1:$

$\implies \sin^2 (6x) + \cos^2 (6x)=1$

$\implies \cos^2 (6x)=1- \sin^2 (6x)$

$\implies 12\displaystyle \int \dfrac{1+\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

Expand:

$\implies 12\displaystyle \int \dfrac{1}{\cos^2(6x)}+\dfrac{\sin(6x)}{\cos^2(6x)} \:\:\text{d}x$

$\textsf{Use the identities }\:\: \sec \theta=\dfrac{1}{\cos \theta} \textsf{ and } \tan\theta=\dfrac{\sin \theta}{\cos \theta}:$

$\implies 12\displaystyle \int \sec^2(6x)+\dfrac{\tan(6x)}{\cos(6x)} \:\:\text{d}x$

$\implies 12\displaystyle \int \sec^2(6x)+\tan(6x)\sec(6x) \:\:\text{d}x$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$

$\boxed{\begin{minipage}{6 cm}\underline{Integrating $ \sec kx \tan kx$}\\\\$\displaystyle \int \sec kx \tan kx\:\text{d}x= \dfrac{1}{k}\sec kx\:\:(+\text{C})$\end{minipage}}$

$\implies 12 \left[\dfrac{1}{6} \tan (6x)+\dfrac{1}{6} \sec (6x) \right]+\text{C}$

Simplify:

$\implies \dfrac{12}{6} \tan (6x)+\dfrac{12}{6} \sec (6x)+\text{C}$

$\implies 2 \tan (6x)+2 \sec (6x)+\text{C}$

Learn more about indefinite integration here:

brainly.com/question/27805589

brainly.com/question/28155016

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