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3 years ago

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You are flying your dragon kite. It's connected to 34yd of string. The kite is directly above the edge of the pond. The edge of

the pond is 29yd from where the kite is tied to the ground. How high is the kite above the edge of pond? ( round answer to nearest tenth)

1 answer:

Naily [24]3 years ago

5 0

Answer:

63

Step-by-step explanation:

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Write an equation (-2, 10), (10, -14)

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$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{-14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-14}-\stackrel{y1}{10}}}{\underset{run} {\underset{x_2}{10}-\underset{x_1}{(-2)}}}\implies \cfrac{-24}{10+2}\implies \cfrac{-24}{12}\implies -2$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{10}=\stackrel{m}{-2}[x-\stackrel{x_1}{(-2)}]\implies y-10=-2(x+2) \\\\\\ y-10=-2x-4\implies y=-2x+6$

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3 years ago

Answer is 60. Plz step by step explanation !! I'm in urgent !! Thank you

Aleks04 [339]

Answer:

60°

Step-by-step explanation:

ABCO is a parallelogram (opposite sides are parallel).

AO is equal to the radius of the circle, so BC must also equal the radius.

CO is equal to the radius of the circle, so AB must also equal the radius.

If we draw a line from O to B, we split the parallelogram into two triangles. Since OB is equal to the radius of the circle, each triangle is an equilateral triangle. Therefore, angle A is 60°.

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