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tamaranim1 [39]
3 years ago
8

A coin dealer is having paper wraps made to go around a stack of coins. The coins have a diameter of 12 mm, and the dealer state

s that the paper should be exactly 36 mm wide, or 3 times the diameter, to go around the coins. Is the coin dealer correct?
Mathematics
2 answers:
tamaranim1 [39]3 years ago
8 0
<span>No, the coin dealer is not correct. The paper would need to be at least 3.14 times the diameter to go around the coin. The circumference of the coin will be diameter times pi, so 12(3.14).</span>
alexgriva [62]3 years ago
3 0

No, not enough paper.

Should be 3.14 instead of 3.

The circumference will be 12 times pi.

(12)(3.14) = 37.68

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Answer:

\text{Triangle 1:}\\\\x=30\sqrt{2},\\\\\text{Triangle 2:}\\x\approx 36.18,\\?\approx12.37

Step-by-step explanation:

*Notes (clarified by the person who asked this question):

-The triangle on the right has a right angle (angle that appears to be a right angle is a right angle)

-The bottom side of the right triangle is marked with a question mark (?)

<u>Triangle 1 (triangle on left):</u>

Special triangles:

In all 45-45-90 triangles, the ratio of the sides is x:x:x\sqrt{2}, where x\sqrt{2} is the hypotenuse of the triangle. Since one of the legs is marked as 30, the hypotenuse must be \boxed{30\sqrt{2}}

It's also possible to use a variety of trigonometry to solve this problem. Basic trig for right triangles is applicable and may be the simplest:

\cos 45^{\circ}=\frac{30}{x},\\x=\frac{30}{\cos 45^{\circ}}=\frac{30}{\frac{\sqrt{2}}{2}}=30\cdot \frac{2}{\sqrt{2}}=\frac{60}{\sqrt{2}}=\frac{60\cdot\sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{60\sqrt{2}}{2}=\boxed{30\sqrt{2}}\\

<u>Triangle 2 (triangle on right):</u>

We can use basic trig for right triangles to set up the following equations:

\cos 20^{\circ}=\frac{34}{x},\\x=\frac{34}{\cos 20^{\circ}}\approx \boxed{36.18},

\tan 20^{\circ}=\frac{?}{34},\\?=34\tan 20^{\circ}\approx \boxed{12.37}

We can verify these answers using the Pythagorean theorem. The Pythagorean theorem states that in all right triangles, the following must be true:

a^2+b^2=c^2, where c is the hypotenuse of the triangle and a and b are two legs of the triangle.

Verify 34^2+(34\tan20^{\circ})^2=\left(\frac{34}{\cos20^{\circ}}\right)^2\:\checkmark

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