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S_A_V [24]
2 years ago
12

Please help me with this question

Mathematics
2 answers:
malfutka [58]2 years ago
5 0
2/9

5/10 *4/9=2/9

ask if you have any more questions
xz_007 [3.2K]2 years ago
3 0

Answer:

2/9

Step-by-step explanation:

there are 5 blue, 4 balck, and 1 green

5+4+1=10

5 blue so 5/10 = 1/2

there are 9 shirts left 4 blue

4/9

1/2 x 4/9 = 4/18 = 2/9

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it would be 17

Step-by-step explanation:

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190/570

Step-by-step explanation:

190/570

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Tatiana [17]

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3 years ago
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.

5 0
3 years ago
Read 2 more answers
Factorise the following quadratic expressions;<br> 6a-18a<br>21x -3x²<br>1/16 - p²<br>25v²-1<br>​
Lubov Fominskaja [6]

Answer:

− 12 a

3 x ( 7 − x )

( 1 4 + p ) ( 1 4 − p )

( 5 v + 1 ) ( 5 v − 1 )

Step-by-step explanation:

I think these are the right answers

3 0
2 years ago
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