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ololo11 [35]
2 years ago
14

Geometry question Need help with #10

Mathematics
1 answer:
Lena [83]2 years ago
6 0

Answer:

x = 6

Step-by-step explanation:

Line segment SU is a bisector of angle <TSV which means it divided the angle into two equal parts:

The measure of <TSV = 62 in this case the measure of <VSU = 62/2 = 31

we can write the following equation to find the value of x

5x + 1 = 31 subtract 1 from both sides

5x = 30 divide both sides by 5

x = 6

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Javier’s Bakery sells donuts in packages of 3 and cookies in packages of 5 today the backers sold the same number of donuts and
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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.

5 0
3 years ago
Read 2 more answers
Hello<br>mere pyaare bhaiyo or bheno​
goldfiish [28.3K]

Cz I secretly enter this site

Non of my family knows

Cz they are over protective lol

Only a cousin know

That's why l can't pay to comment

It's sad life bro

5 0
2 years ago
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