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2 years ago

Someone help me with maths questions pls about pythagoras, ill give u brainliest

Mathematics

1 answer:

Marat540 [252]2 years ago

6 0

Answer:

11.183 cm and 22.366 cm

Step-by-step explanation:

Lets call the 25 cm hypotenuse C and the other sides A and B.

B=2A

The Pythagorean theorem states that C^2=A^2+B^2 so:

25^2=A^2+B^2

substitute for B

25^2=A^2+(2A)^2

625=A^2+4A^2

625=5A^2

125=A^2

A=11.183 cm

B=22.366 cm

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3 years ago

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miv72 [106K]

Hint :-

Break the given sequence into two parts .
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Notice the terms beginning from 2nd term , $4,7,10,13$ . Next term is obtained by adding 3 to the previous term .

Solution :- 

We need to find out the sum of 50 terms of the given sequence . After splitting the given sequence ,

$\implies S_1 = \dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8}$ .

We can see that this is in Geometric Progression where 1/2 is the common ratio . Calculating the sum of 25 terms , we have ,

$\implies S_1 = a\dfrac{1-r^n}{1-r} \\\\\implies S_1 = \dfrac{1}{2}\left[ \dfrac{1-\bigg(\dfrac{1}{2}\bigg)^{25}}{1-\dfrac{1}{2}}\right]$

Notice the term $\dfrac{1}{2^{25}}$ will be too small , so we can neglect it and take its approximation as 0 .

$\implies S_1\approx \cancel{ \dfrac{1}{2} } \left[ \dfrac{1-0}{\cancel{\dfrac{1}{2} }}\right]$

$\\\implies \boxed{ S_1 \approx 1 }$

$\rule{200}2$

Now the second sequence is in Arithmetic Progression , with common difference = 3 .

$\implies S_2=\dfrac{n}{2}[2a + (n-1)d]$

Substitute ,

$\implies S_2=\dfrac{25}{2}[2(4) + (25-1)3] =\boxed{ 908}$

Hence sum = 908 + 1 = 909

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3 years ago

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andrezito [222]

Answer:

yes

Step-by-step explanation:

0.045 is greater than 0.00195

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