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3 years ago

Which segment is parallel to BC

Mathematics

1 answer:

adell [148]3 years ago

3 0

Line DF is parallel

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Step-by-step explanation:

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=39000

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3 years ago

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The choice A ;

$\frac{1}{ {7}^{8} }$

Step-by-step explanation:

$\frac{ {7}^{ - 6} }{ {7}^{2} } = \frac{ {7}^{ - 6 - 2} }{1} = {7}^{ - 8} = \frac{1}{ {7}^{8} } \\ \\ \\ or. \: \frac{ {7}^{ - 6} }{ {7}^{2} } = \frac{1}{ {7}^{ 2 + 6} } = \frac{1}{ {7}^{8} }$

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2 years ago

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If dy dx equals cosine squared of the quantity pi times y over 4 and y = 1 when x = 0, then find the value of x when y = 3.

Afina-wow [57]

Answer: $x = -\frac{8}{\pi}$

Explanation: Note that

$\frac{dy}{dx} = \cos^2 \left ( \frac{\pi y}{4} \right )
\\
\\ \frac{dy}{\cos^2 \left ( \frac{\pi y}{4} \right )} = dx
\\
\\ \int{\frac{dy}{\cos^2 \left ( \frac{\pi y}{4} \right )}} = \int dx
\\
\\ \boxed{x = \int{\sec^2 \left ( \frac{\pi y}{4} \right )dy}}$

To evaluate the integral in the boxed equation, let $u = \frac{\pi y}{4}$ . Then,

$du = \frac{\pi}{4} dy 
\\ \Rightarrow \boxed{dy = \frac{4}{\pi}du}$

So,

$x = \int{\sec^2 \left ( \frac{\pi y}{4} \right )dy}
\\
\\ = \int{\sec^2 u \left ( \frac{4}{\pi}du \right )}
\\
\\ = \frac{4}{\pi}\int{\sec^2 u du}
\\
\\ = \frac{4}{\pi} \tan u + C
\\
\\ \boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) + C}\text{ (1)}$

Since y = 1 when x =0, equation (1) becomes

$0 = \frac{4}{\pi} \tan \left ( \frac{\pi (1)}{4} \right ) + C 
\\ 
\\ 0 = \frac{4}{\pi} (1) + C 
\\
\\ \frac{4}{\pi} + C = 0
\\
\\ \boxed{C = -\frac{4}{\pi}}$

With the value of C, equation (1) becomes

$\boxed{x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} }$

Hence, if y = 3,

$x = \frac{4}{\pi} \tan \left ( \frac{\pi y}{4} \right ) -\frac{4}{\pi} 
\\
\\ x = \frac{4}{\pi} \tan \left ( \frac{\pi (3)}{4} \right ) -\frac{4}{\pi} 
\\
\\ x = \frac{4}{\pi} (-1) -\frac{4}{\pi} 
\\
\\ \boxed{x = -\frac{8}{\pi}}$

6 0

3 years ago