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Liono4ka [1.6K]
1 year ago
11

If four is half of five and two thirds of six is nine. what is thirty-two?

Mathematics
1 answer:
valkas [14]1 year ago
5 0
Four is IV which is half of five. IX is nine which is 2 thirds of 6 so... thirty-two is 1 5/6
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1 year ago
Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes
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Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

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Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

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At 5 pm:

Distance traveled by boat 1 = 33 + 2\times 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 \times 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled \triangle AOB.

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = \dfrac{Total\ distance}{Total\ time}

\Rightarrow \dfrac{74.28}{5} = 14.86\ knots

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Irina-Kira [14]
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c^{2} = a^{2} + b^{2} - 2abcosC

Rearranging this equation, we can find the value ∠C as shown below.

\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}
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We can apply the same reasoning for finding the value of ∠B as shown.

B = cos^{-1} (\frac{a^{2}+c^{2}-b^{2}}{2ac})

Plugging in the values of the sides (see image attached) from the given. It will now be straightforward to compute for ∠B and ∠C.

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7 0
2 years ago
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