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Andreyy89
3 years ago
7

when the length of a rectangle is X inches and the width is y inches, the perimeter is 30 inches. if the length is tripled and t

he width is quadrupled the perimeter becomes 102 inches. find the length and width of the original rectangle
Mathematics
1 answer:
Elena L [17]3 years ago
5 0

P1=2x+2y=30

after modify them

P2=2*3x+2*4y=102

     6x+8y=102

      2x+2y=30      / * (-3)

      -6x-6y= -90

       6x+8y=102

we must add the 2 expressions

         / 8y-6y=102-90

            2y=12

           y=12:2

          y=6

       2*x+2*y=30

       2x+2*6=30

       2x+12=30

      2x=30-12

     2x=18

      x=18:2

      x=9



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Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.
const2013 [10]

Answer:

a.

T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) |  =   (-cos(t^2) , -sin(t^2))

b.

T(t) =r'(t)/|r'(t)| =  (2t/ \sqrt{4t^2   + 36} , -6/\sqrt{4t^2   + 36},0)

N(t) =T(t)/|T'(t)| =  (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)

Step-by-step explanation:

Remember that for any curve      r(t)  

The tangent vector is given by

T(t) = \frac{r'(t) }{| r'(t)| }

And the normal vector is given by

N(t) = \frac{T'(t)}{|T'(t)|}

a.

For this case, using the chain rule

r'(t) = (  -10*2tsin(t^2) ,   102t cos(t^2)   )\\

And also remember that

|r'(t)| = \sqrt{(-10*2tsin(t^2))^2  +  ( 10*2t cos(t^2) )^2} \\\\       = \sqrt{400 t^2*(  sin(t^2)^2  +  cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t

Therefore

T(t) = r'(t) / |r'(t) | =  (  -10*2tsin(t^2) ,   10*2t cos(t^2)   )/ 20t\\\\ = (  -10*2tsin(t^2)/ 20t ,   10*2t cos(t^2) / 20t  )\\= ( -sin(t^2), cos(t^2) )

Similarly, using the quotient rule and the chain rule

T'(t) = ( -2t cos(t^2) , -2t sin(t^2))

And also

|T'(t)| = \sqrt{  ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t

Therefore

N(t) = T'(t) / |T'(t) |  =   (-cos(t^2) , -sin(t^2))

Notice that

1.   |N(t)| = |T(t) | = \sqrt{ cos(t^2)^2  + sin(t^2)^2 } = \sqrt{1} =  1

2.   N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0

b.

Simlarly

r'(t) = (2t,-6,0) \\

and

|r'(t)| = \sqrt{(2t)^2   + 6^2} = \sqrt{4t^2   + 36}

Therefore

T(t) =r'(t)/|r'(t)| =  (2t/ \sqrt{4t^2   + 36} , -6/\sqrt{4t^2   + 36},0)

Then

T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)

and also

|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 +   ( (3 t)/(9 + t^2)^{3/2})^2  +  0^2 }\\= 3/(t^2 + 9 )

And since

N(t) =T(t)/|T'(t)| =  (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)

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Dima020 [189]
The answer to the first one is 72.

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laiz [17]
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In the diagram, point D divides line segment AB in the ratio of 5:3. If line segment AC is vertical and line segment CD is horiz
BabaBlast [244]

Answer:

C (2, -1)

Step-by-step explanation:

C has the same x coordinate as A = 2

that eliminates A and B.

and the y value of C has to be closer to the y value of B than to the y value of A (due to the 5:3 ratio).

so, without spending unnecessary effort on actual calculation, we can directly derive the correct answer to be C.

but to show in numbers, the "coordinate triangle" of the difference between D and B is a similar triangle to ACD. and so, the ratio of one side pair applies then to all other sides too. including the individual coordinate differences.

so, also the y difference of A to D relates to the y difference between D and B in the same ratio of 5:3 of the overall y difference of A to B.

the y difference A to B is 2 - -6 = 8

the 5/3 split of 8 units is actually 5 and 3.

so, the y coordinate of D is -6 + 5 = -1

and C must have the same y coordinate.

therefore, it is proven that answer C (2, -1) is the right answer.

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Answer:

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Step-by-step explanation:

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7 0
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