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Yuki888 [10]
3 years ago
8

"Graph y - 2 = 2/3 (x + 4)"

Mathematics
1 answer:
Firdavs [7]3 years ago
6 0

It would look like this

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I need help with this question
Anna11 [10]
J.WYIBVNJVJNDVBBNUVHUGHTNBJTHNIL
3 0
3 years ago
Movies Plus charges its customers a $10 monthly service fee plus $2 for each movie the customer rents. Movies For Less
Vinvika [58]

Answer:

A

Step-by-step explanation:

10x3=30

15x2=30+10=40

15x3=45 movies for less

7 0
3 years ago
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
You buy 3.18 pounds of peaches​, 1.45 pounds of apples​, and 2.06 pounds of pears. What is your total​ bill?
Mrrafil [7]

Answer:

6.69 is the total amount of pounds

6 0
3 years ago
Read 2 more answers
NEED HELP FAST WILL GIVE BRAINLIEST
Olin [163]
The answer is A, as I calculated the difference using Pythagora's Theorem. Which I then did 7.6-6.3= 1.29122, which rounds to 1.0 cm^2
7 0
3 years ago
Read 2 more answers
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