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garik1379 [7]
3 years ago
8

PLEASE HELP!! 50 points and brainliest!!!

Mathematics
2 answers:
Kitty [74]3 years ago
8 0
EQUIVALENT WOULD BE THE ANSWER 

ivolga24 [154]3 years ago
7 0
The correct answer would be B. Equivalent 

Since the line is dependent at a straight line, it would make it equivalent to being the same. 

Hope this helps! 

<em>~ ShadowXReaper069</em>




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You should divide 280 into 3 which would equal 9
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Please help me ill mark brainliest
Ivanshal [37]

Answer:

its C and 4.7x10 -5 was right

Step-by-step explanation:

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Can someone help me out on this plz?
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c.) 36cm

Step-by-step explanation:

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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
Can someone please help me answer this?
sammy [17]

Answer:

Not clear enough

Step-by-step explanation:

please resend the question and make it more clear and readable

8 0
3 years ago
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