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Fofino [41]
3 years ago
11

How do you solve y=7x+35

Mathematics
1 answer:
12345 [234]3 years ago
6 0

Answer:

with ur head..........

Step-by-step explanation:

n/a

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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)

6 0
3 years ago
What is the measurement of angle POR
IgorLugansk [536]

Let us observe the given figure,

When two lines intersect each other, the angles opposite to each other are Vertically Opposite Angles. Vertically opposite angles are always equal in measure.

As, we can observe that the given lines intersect each other, and they form vertically opposite angles as \angle POR , \angle SOQ and  \angle POS , \angle ROQ

Therefore, \angle POR = \angle SOQ

Substituting the given measures of the angles, we get

91-x^\circ = x+ 7^\circ

91-7 = x + x

84 = 2x

x = \frac{84}{2}

So, x = 42^\circ

Since, the measure of angle POR = (91-x)^\circ

= (91-42)^\circ

= 49^\circ

Therefore, the measure of angle POR is 49 degrees.

8 0
3 years ago
Can Anyone Help me with this I believe I got the first one correct If not correct me please
olga_2 [115]

Answer:

the second one I think

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
A value meal package at Ron's Subs consists of a drink, a sandwich, and a bag of chips. There are 66 types of drinks to choose f
Ludmilka [50]

36 different value meal packages are possible

Step-by-step explanation:

To answer this question, multiply all given numbers together.

4*3*3

12*3

36

4 0
3 years ago
Plz HELP will give extra points
jekas [21]
The mode is 44
The range is 3
The median is 44
The mean is 44
4 0
3 years ago
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