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babymother [125]
3 years ago
10

How did British scientists test Einstein theory of relativity

Mathematics
1 answer:
Stolb23 [73]3 years ago
8 0

Answer: Arthur Stanley Eddington's 1919 expedition confirmed Einstein's prediction for the deflection of light by the Sun during the total solar eclipse of 29 May 1919 which helped to cement the status of general relativity as a true theory. Since then many observations have confirmed the correctness of general relativity.

Hope this helps. Also gimme brainliest.

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Pls help asap I will give brainerlist plus 15 points
xxMikexx [17]

Answer:

52

Step-by-step explanation:

6^2= 36

4^2= 16

36+16=52

3 0
3 years ago
Plz no links and answers asap
JulijaS [17]

Answer:

10 is the evaluated answer.

4 0
3 years ago
Read 2 more answers
Cos(−θ)=√3/3, sinθ<0
Delicious77 [7]

Answer:

\sin\theta=-\frac{\sqrt{6}}{3}


Step-by-step explanation:

The given trigonometric equation is \cos(-\theta)=\frac{\sqrt{3} }{3}.

We can either use the Pythagorean identity or the right angle triangle  to solve for \sin\theta.

According to the Pythagorean identity,

\cos^2\theta+\sin^2\theta=1


Recall that, the cosine function is an even function, therefore

\cos(-\theta)=\cos(\theta)


\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}.

We substitute this value in to the above Pythagorean identity to get;


(\frac{\sqrt{3}}{3})^2+\sin^2\theta=1


\Rightarrow \frac{3}{9}+\sin^2\theta=1


\Rightarrow \sin^2\theta=1-\frac{3}{9}


\Rightarrow \sin^2\theta=\frac{6}{9}


\Rightarrow \sin\theta=\pm \sqrt{\frac{6}{9}}


\Rightarrow \sin\theta=\pm \frac{\sqrt{6}}{3}


But we were given that,

\sin\theta\:, so we choose the negative value.

\Rightarrow \sin\theta=-\frac{\sqrt{6}}{3}


The correct answer is B







7 0
3 years ago
Please help ill give brainliest<br>​
Paul [167]
1.c
2.d
3.b
4.a
Im sure about 1 and 4 but a little hesitant on 2 and 3 but good luck pretty sure it’s right
3 0
3 years ago
Completing the square
kogti [31]
Please provide a square, thank you !!!
8 0
3 years ago
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