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3 years ago

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A Ladder That Is 5 meters in length is resting on a branch of a tree and the base of the ladder is 3 meters from the tree on the

level ground. How High Up is the branch on which the ladder is resting

1 answer:

Stells [14]3 years ago

5 0

Use the Pythagoras theorem:-

height = sqrt(5^2 - 3^2 ) = sqrt16 = 4 metres Answer.

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Jermey swims 10 laps in 15 minutes. How many minutes per lap is that?

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Answer:

1,5

Step-by-step explanation:

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Step-by-step explanation:

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Suppose each edge of the cube shown in the figure is 10 inches long. Find the sine and cosine of the angle formed by diagonals D

tatiyna

Check the picture below.

$sin(EDG )=\cfrac{\stackrel{opposite}{10}}{\underset{hypotenuse}{10\sqrt{3}}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\implies sin(EDG )=\cfrac{1}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{sin(EDG )=\cfrac{\sqrt{3}}{\sqrt{3^2}}\implies sin(EDG )=\cfrac{\sqrt{3}}{3}} \\\\[-0.35em] ~\dotfill$

$cos(EDG )=\cfrac{\stackrel{adjacent}{10\sqrt{2}}}{\underset{hypotenuse}{10\sqrt{3}}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\implies cos(EDG )=\cfrac{\sqrt{2}}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{cos(EDG )=\cfrac{\sqrt{6}}{\sqrt{3^2}}\implies cos(EDG )=\cfrac{\sqrt{6}}{3}}$

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2 years ago

Bill rented a truck for one day. There was a base fee of $20.99, and there was an additional charge of 88 cents for each mile dr

lana66690 [7]

Answer:

262 miles

Step-by-step explanation:

251.55 - 20.99 = 230.56

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