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3 years ago

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Mathematics

1 answer:

Alika [10]3 years ago

3 0

Answer:

10 cats were at the shelter on wednesday

Step-by-step explanation:

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Option A: 40 trash bags for $6.29 Option B: 24 trash bags for $3.89

mr Goodwill [35]

Option B is a better option.

Step by step:

I divided 40/6.29 and got about 6.3.

I divided 24/3.89 and got about 6.1.

Which is less, 6.3 or 6.1?
If you said 6.1 you are correct, so option B has a less cost, making it a better option.

8 0

2 years ago

What is the value of the expression below when y=2 and z=8<br> 8y-z

MrRissso [65]

Answer:

Step-by-step explanation:

y = 2

z = 8

8y - z

= 8 ( 2 ) - 8

= 16 - 8

= 8

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:

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

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

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

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

WILL AWARD BRAINLIEST (Find the measure of the complement and supplement)

melisa1 [442]

Answer:

The complement is 53 and the supplement is 143

Step-by-step explanation:

Two complementary angles add to 90

PQR and x = 90

37+x = 90

Subtract 37 from each side

37-37 +x =90-37

x = 53

The complement is 53

Two supplementary angles add to 180 degrees

PQR and x = 180

37+x = 180

Subtract 37 from each side

37-37 +x= 180-37

x = 143

The supplement is 143

8 0

3 years ago

Simplify <br> (5x2 + 3x + 4) − (2x2 − 6x + 3)

dangina [55]

Answer is <span>7x^2-3x+7
hope this helps</span>

5 0

3 years ago

Read 2 more answers