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3 years ago

a painting including the frame measures 71 cm by 75 cm How many centimeters of framing were needed to frame the painting?

Mathematics

1 answer:

san4es73 [151]3 years ago

7 0

I think 292cm, because it is asking for the perimeter

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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:

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

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

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

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

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

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

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

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

6 0

3 years ago

In triangle abc shown below side ab is 6 and side ac is 4

kaheart [24]

Question is Incomplete, Complete question is given below:

In Triangle ABC shown below, side AB is 6 and side AC is 4.

Which statement is needed to prove that segment DE is parallel to segment BC and half its length?

Answer

Segment AD is 3 and segment AE is 2.

Segment AD is 3 and segment AE is 4.

Segment AD is 12 and segment AE is 4.

Segment AD is 12 and segment AE is 8.

Answer:

Segment AD is 3 and segment AE is 2.

Step-by-step explanation:

Given:

side AB = 6

side AC = 4

Now we need to prove that segment DE is parallel to segment BC and half its length.

Solution:

Now AD + DB = AB also AE + EC = AC

DB = AB - AD also EC = AC - AE

Now we take first option Segment AD is 3 and segment AE is 2.

Substituting we get;

DB = 6-3 = 3 also EC = 4-2 =2

From above we can say that;

AD = DB and EC = AE

So we can say that segment DE bisects Segment AB and AC equally.

Hence From Midpoint theorem which states that;

"The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side."

Hence Proved.

7 0

3 years ago

Which situation can be represented by the equation below?

pav-90 [236]

Answer:

Where is the equation?

7 0

3 years ago

Brayden and gavin were playing touch football against cole and freddy touchdowns were worth 7 points.brayden and gavin scored 7

larisa [96]

Answer: 14 points or 2 touchdowns

Step-by-step explanation:

7 touchdowns times 7 points = 49 points scored by brayden and gavin

9 touchdowns times 7 points - 63 points scored by cole and freddy

63 - 49 = 14

14 points = 2 touchdowns

8 0

3 years ago

What is the slope of the line that passes through (-4, 6) and (-6, 20)?

Hatshy [7]

Answer:

-7

Step-by-step explanation:

Since we have two points, we can use the slope formula

m = ( y2-y1)/(x2-x1)

= ( 20-6)/(-6- -4)

= ( 20-6)/ ( -6+4)

= 14/-2

= -7

3 0

3 years ago

a painting including the frame measures 71 cm by 75 cm How many centimeters of framing were needed to frame the​ painting?

a painting including the frame measures 71 cm by 75 cm How many centimeters of framing were needed to frame the painting?