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timama [110]
3 years ago
12

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

=[4,0,-3]
Mathematics
1 answer:
SashulF [63]3 years ago
3 0

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

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Can I get some help, it is due in 15 minutes and I need help. Thanks, any help appreciated
koban [17]

Well, I'm way past the 15 min mark, but here's how to do the question.


With this, you will need to use the distance formula, \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, on XY, YZ, and ZX.



XY: \sqrt{(3-1)^2+(1-6)^2}


Firstly, solve inside the parentheses: \sqrt{(2)^2+(-5)^2}


Next, solve the exponents: \sqrt{4+25}


Next, solve the addition, and XY's distance will be √29



(The process is the same with the other 2 sides, so I'll go through them real quickly)


YZ:

\sqrt{(6-3)^2+(3-1)^2}\\ \sqrt{(3)^2+(2)^2}\\ \sqrt{9+4}\\ \sqrt{13}



ZX:

\sqrt{(1-6)^2+(6-3)^2}\\ \sqrt{(-5)^2+(3)^2}\\ \sqrt{25+9}\\ \sqrt{34}



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8 0
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A golfer keeps track of his score for playing nine holes of golf​ (half a normal golf​ round). His mean score is 8080 with a sta
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Answer:

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and 11\sqrt2 respectively.

Step-by-step explanation:

Given that,

For the first 9 holes X:

E(X) = 80

SD(X)=13

For the second 9 holes Y:

E(Y) = 80

SD(Y)=13

For the sum W=X+Y, the following properties holds for means , variance and standard deviation :

E(W)=E(X)+E(Y)

     and

V(W)=V(X)+V(Y)

⇒SD²(W)=SD²(X)+SD²(Y)        [ Variance = (standard deviation)²]

\Rightarrow SD(W)=\sqrt{SD^2(X)+SD^2(Y)}

∴E(W)=E(X)+E(Y) = 80 +80=160

            and

∴SD(W)=\sqrt{SD^2(X)+SD^2(Y)}

              =\sqrt{11^2+11^2}

               =\sqrt{2.11^2}

               =11\sqrt2

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and 11\sqrt2 respectively.

                                               

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