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sveticcg [70]
3 years ago
11

Find the GCF (Greatest Common Factor) of 64 and 144

Mathematics
2 answers:
Ipatiy [6.2K]3 years ago
6 0
The greatest common factor is 64
valkas [14]3 years ago
3 0

Answer: 16

Step-by-step explanation:

if we were to list all the factors of 64 and 144

64= 1,2,4,8,8,16,32,64

144= 1,2,3,4,6,8,9,12,12,16,18,24,36,48,72,144

therefore the GCF of 64 and 144 is 16

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Answer:

<em>Proof in the explanation</em>

Step-by-step explanation:

<u>Trigonometric Equalities</u>

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y

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\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}  

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\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})

And modify the first variable:

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Now with the second variable

\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}

Knowing that

sin^2x+cos^2x=1

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\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}

\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}

\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}

Computing

15-6\sqrt{6}=(3-\sqrt{6})^2

Then

\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}

Now we replace all in the first equality:

\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}

\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}

\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6

Thus, proven  

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