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Kruka [31]
3 years ago
7

What is 8 1/8 divided by 5/6 as a fraction in simplest form

Mathematics
1 answer:
svetoff [14.1K]3 years ago
7 0

Answer:

3.9 but in fraction form it would be 3 9/10

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In triangle​ abc, the measure of angle b is 32 degrees more than three times the measure of angle
Yuki888 [10]
Define x:
Let angle a be x.
angle a = x
angle b = 3x + 32
angle c = x + 58

Find x:
Angles in a triangle add up to 180°:
x + 3x + 32 + x + 58 = 180

Combine like terms:
5x + 90 = 180

Take away 90 from both sides:
5x = 90

Divide by 5 on both sides:
x = 18°

Find the angles:
angle a = x = 18°
angle b = 3x + 32 = 3(18) + 32 = 86°
angle c = x + 58 = 18 + 58 = 76°

Answer: The angles are 18°, 86° and 76°.
7 0
3 years ago
2. Afton made a chicken dish for dinner. Sheadded a 10-ounce package of vegetablesand a 14-ounce package of rice to 40 ouncesof
Bad White [126]
First, we need to add up all the quantities in ounces. 10 + 14 + 40 = 64. One pound = 16 ounces. 64 / 16 = 4, so the total weight in pounds is 4 lbs.
4 0
3 years ago
List the domain and range if the relation {(0,2),(4,3),(5,5),(4,7)}
lyudmila [28]
(x,y)

domain is inputs (normally x or first number)
range is outputs from the given domain (normally y or second number)

domain is just all the first numbers
range is all the second numbers
ignore repeats


domain: {0,4,5}
range: {2,3,5,7}
5 0
3 years ago
Read 2 more answers
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
Plese help me its timed
Nadusha1986 [10]

Answer:

(3,2) I think

Step-by-step explanation:

X= 3

Y=2

3 +2 =5

3-2=1

3 0
2 years ago
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