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3 years ago

Chords and arcs. Can someone please help me with this and explain???20 points

Mathematics

1 answer:

anyanavicka [17]3 years ago

5 0

Answer:

Step-by-step explanation:

So the line segment CD is 12.7 and half that is 6.35. I wanted this 6.35 so I can look at the right triangle there and find the angle there near the center. This will only be half the answer. So I will need to double that to find the measure of arc CD.

Anyways looking at angle near center in the right triangle we have the opposite measurement, 6.35, given and the hypotenuse measurement, 9.06, given. So we will use sine.

sin(u)=6.35/9.06

u=arcsin(6.35/9.06)

u=44.5 degrees

u represented the angle inside that right triangle near the center.

So to get angle COD we have to double that which is 89 degrees.

So the arc measure of CD is 89.

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3 years ago

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Answer:

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3 years ago

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3 years ago

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meriva

Answer:

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3 years ago

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$\bf \lim\limits_{x\to 0}\ \cfrac{sin^3(x)}{5x^3}-4
\\ \quad \\\\ \cfrac{sin^3(x)}{5x^3}-4\implies \cfrac{[sin(x)]^3}{5x^3}-4
\\ \quad \\
 

\cfrac{[sin(x)]^3}{x^3}\cdot \cfrac{1}{5}-4
\\ \quad \\
thus
\\ \quad \\
\lim\limits_{x\to 0}\cfrac{[sin(x)]^3}{x^3}\cdot \lim\limits_{x\to 0}\cfrac{1}{5}-4\qquad \boxed{recall \qquad \lim\limits_{x\to 0} \cfrac{sin(x)}{x}\implies 1}
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1\cdot \cfrac{1}{5}-4$

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3 years ago