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2 years ago

HELP PLEASE LAST ONE

Mathematics

1 answer:

S_A_V [24]2 years ago

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Mdkdgsmsmsgsbjslsgsvsvsvsjs

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How much meters are in how many cm

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phoebe needs to use 3/4 cup to make one batch of cookies how many batches can she make with 12 cups of sugar?

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Thank you for posting your question here at brainly. Feel free to ask more questions.

The best and most correct answer among the choices provided by the question is A. Between 0.05 and 0.01 .
 
Hope my answer would be a great help for you.

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Determine whether the given vectors are orthogonal, parallel or neither. (a) u=[-3,9,6], v=[4,-12,-8,], (b) u=[1,-1,2] v=[2,-1,1

nevsk [136]

Answer:

a) $u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

b) $u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

$cos \theta = \frac{uv}{|u| |v|}$

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

c) $u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

Step-by-step explanation:

For each 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.

Part a

u=[-3,9,6], v=[4,-12,-8,]

The dot product on this case is:

$u v= (-3)*(4) + (9)*(-12)+ (6)*(-8)=-168$

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:

$|u|= \sqrt{(-3)^2 +(9)^2 +(6)^2}=\sqrt{126}$

$|v| =\sqrt{(4)^2 +(-12)^2 +(-8)^2}=\sqrt{224}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{-168}{\sqrt{126} \sqrt{224}})=cos^{-1} (-1) = \pi$

Since the angle between the two vectors is 180 degrees we can conclude that are parallel

Part b

u=[1,-1,2] v=[2,-1,1]

The dot product on this case is:

$u v= (1)*(2) + (-1)*(-1)+ (2)*(1)=5$

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:

$|u|= \sqrt{(1)^2 +(-1)^2 +(2)^2}=\sqrt{6}$

$|v| =\sqrt{(2)^2 +(-1)^2 +(1)^2}=\sqrt{6}$

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

$cos \theta = \frac{uv}{|u| |v|}$

And the angle is given by:

$\theta = cos^{-1} (\frac{uv}{|u| |v|})$

If we replace we got:

$\theta = cos^{-1} (\frac{5}{\sqrt{6} \sqrt{6}})=cos^{-1} (\frac{5}{6}) = 33.557$

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

Part c

u=[a,b,c] v=[-b,a,0]

The dot product on this case is:

$u v= (a)*(-b) + (b)*(a)+ (c)*(0)=-ab +ba +0 = -ab+ab =0$

Since the dot product is equal to zero then the two vectors are orthogonal.

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23% would be the answer

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