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Dima020 [189]
3 years ago
13

Melinda rents paddle boats down at the pier. She uses 5.5 feet of nylon rope to tie off each boat. She started with 80 feet of r

ope. After securing all the boats, she was left with more than 3 feet of rope. How many boats did Melinda tie off?
A.
at most 16
B.
at most 26
C.
less than 14
D.
more than 16
Mathematics
1 answer:
Alina [70]3 years ago
7 0
B. At most 26 is the answer
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A bug dug 60 inches straight down below ground and then continued digging 144 inches parallel to the ground. Finally, he realize
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156 inches is the shortest distance to dig diagonally through dirt to get to his family.

<u>Step-by-step explanation:</u>

  • A bug dug 60 inches straight down below ground.
  • Then continued digging 144 inches parallel to the ground.

This forms the angle 90° between the two lines.

After he realized to turn back., the shortest distance is to dig diagonally.

This forms the hypotenuse side of the triangle.

Therefore, the given data can be assumed as a right angled triangle.

<u>To find the diagonal path (hypotenuse) :</u>

Use the Pythagorean triplets for right angle triangle,

We know that the triplet is (5,12,13).

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Therefore, the third side hypotenuse must also be a multiple of 12.

From the triplets (5,12,13), the hypotenuse is 13 and it is multiplied by 12.

Diagonal path ⇒ 13 × 12 = 156  inches.

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3 years ago
Ishaan is 3 times as old as Chris and is also 14 years older than Chris. How old is ishaan?
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Katyanochek1 [597]

Write tan in terms of sin and cos.

\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}

Recall that

\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1

Rewrite and expand the given limand as the product

\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)

Then using the known limit above, it follows that

\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}

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