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yulyashka [42]
3 years ago
13

What is the length of a hypotenuse of a triangle if each of its legs is 4 units

Mathematics
2 answers:
azamat3 years ago
8 0

Answer:

\boxed{c = 5.7 units}

Step-by-step explanation:

<u><em>Using Pythagorean Theorem:</em></u>

=> c^2 = a^2+b^2

Where c is hypotenuse, a is base and b is perpendicular and ( a, b = 4)

=> c^2 = 4^2+4^2

=> c^2 = 16+16

=> c^2 = 32

Taking sqrt on both sides

=> c = 5.7 units

alukav5142 [94]3 years ago
6 0

Answer:

<h2>5.65 units</h2>

Step-by-step explanation:

Given,

Base ( b ) = 4 units

Perpendicular ( p ) = 4 units

Hypotenuse ( h ) = ?

Now,

Using Pythagoras theorem to find length of hypotenuse:

{h}^{2}   =  {p}^{2}  +  {b}^{2}

Plugging the values

{h}^{2}  =  {4}^{2}  +  {4}^{2}

Evaluate the power

{h}^{2}  = 16 + 16

Calculate the sum

{h}^{2}  = 32

h =  \sqrt{32}

h = 5.65 \: units

Hope this helps..

Best regards !!

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nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

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Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

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2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

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3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

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What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

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