What is the radius of the sun?

Mathematics

1 answer:

Harrizon [31]3 years ago

5 0

432,288 miles.
Google something like this lol

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find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B

SashulF [63]

Answer:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

$a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10$

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

$|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3$

$|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5$

And finally we can calculate the angle between the vectors like this:

$cos \theta = \frac{ab}{|a| |b|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{ab}{|a| |b|})$

If we replace we got:

$\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190$

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately $\theta \approx 48$

3 0

3 years ago

Please help me with this??

Ludmilka [50]

Answer:

c, d and f are rational numbers

Step-by-step explanation:

$c) 3\sqrt{5}*2\sqrt{5}=3*2*\sqrt{5*5} = 3*2*5 = 30\\\\\\d) \dfrac{3\sqrt{5}}{2\sqrt{5}}= \dfrac{3}{2}\\\\\\f) \dfrac{2\sqrt{5}}{3\sqrt{5}}= \dfrac{2}{5}$