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svet-max [94.6K]
2 years ago
9

Regular pentagonal tiles and triangular tiles are arranged in the pattern shown. The pentagonal tiles are all the same size and

shape and triangular tiles are all the same size and shape. Find the angle measures of the triangular tiles.

Mathematics
1 answer:
professor190 [17]2 years ago
7 0
First of all you have to find the missing measurements. The actual measurements for the angles in the hexagon are not given, but they give you an expression. You have to solve for x first so that you can plug it in and find the angle measurement. You have to equal the two sides that are given to you like this: 20x+48=33x+9. You solve for x and then plug it into each angle measurement. This should give you 108. Since it is a regular hexagon all of the sides are equal. If you look at the angle at the top of the hexagon you'll see two triangles and the angle. Since it lies on a straight line, it is all equal to 180. You already have the angle measurement of the hexagon and are missing the triangles. So 180-108=72. 72 is the missing part of the angle. You divide this by 2 in order to find each triangle angle measurements. the answer is 36 degrees. 
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.072 divided by 345.00 and round to the nearest tenth. Thank you!
alexandr402 [8]
0.072 ÷ 345

= \frac{0.072}{345} = 0.000208695652174 ≈ 0.00020
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we have f(x)=x^3 which is a cubic function.

and =(x-1)^3+4.

from f(x) and g(x) we know that both are cubic and g(x) has shrink of 1 and up by 4 units in x axis .

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part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.​
Nat2105 [25]

Answer:  \bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}

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

Pythagorean Theorem is: a² + b² = c²   , <em>where "c" is the hypotenuse</em>

<em />

9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}

Note: 4² + (8√2)² = hypotenuse²   →   hypotenuse = 12

10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}

Note: 12² + opposite² = 20²   →   opposite = 16

11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}

Note: adjacent² + 5² = 6²   →   adjacent = √11

12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}

Note: adjacent² + 7² = (13√2)²   →   adjacent = 17

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2 years ago
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as far as I can tell, is just a matter of going around the circle many or infinite times around.

so 6,31° is the first point, the next point will be one-go-around, 6, 31+360 => 6, 391°

then the next will be 6, 391+360 => 6, 751° and so on.

so we can say is (6, 31° ±360°n), n ∈ ℤ.

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The equation is 10x+20y=150 
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