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3 years ago

Pllllllzzzzzz gelpo me aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Mathematics

2 answers:

Savatey [412]3 years ago

7 0

Answer:

FE = 10, DE = 24

Step-by-step explanation:

We can use ratios to solve

13 5

----- = ----------

26 FE

Using cross products

13 FE = 26*5

Divide each side by 13

13 FE/13 = 26*5/13

FE = 10

13 12

----- = ----------

26 DE

Using cross products

13DE = 12*26

Divide each side by 13

13 DE/13 = 26*12/13

DE = 24

Ilia_Sergeevich [38]3 years ago

3 0

Answer:

D. DE = 24 yd, EF = 10 yd

Step-by-step explanation:

Solve for the unknown angles by setting up common ratios and cross multiplying. You can use this method because they are similar triangles.

For example, DE can be found by setting the ratios of AC/AB equal to the ratio of DF/DE.

13/12 = 26/x

Simply solve for x!

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Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

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Answer:

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

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

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The tangent vector is given by

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

And the normal vector is given by

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

For this case, using the chain rule

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

And also remember that

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

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

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

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

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and

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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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