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2 years ago

Don’t be all up on me like that

Mathematics

2 answers:

Lelechka [254]2 years ago

7 0

Sucks </3

HSGYAUSHBJASKDUGAYHD

anyways

goodluck

Anit [1.1K]2 years ago

5 0

Answer:

for you explanation just put i chose 6 to the third power because that is the only answer that does not equal 64

Step-by-step explanation:

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The dashed curve on the graph shows the initial function . What is the equation of the solid curve?

Jet001 [13]

Answer:

f(x) = sqrt(x + 6)

y = sqrt(x + 6)

x = sqrt(y + 6)

Stx^2 = y + 6

x^2 - 6 = y

red line: f(x) = sqrt(x + 6)

green line: f^-1(x) = x^2 - 6

7 0

3 years ago

(x-4) + 34 = -4x I need this answer

mote1985 [20]

The answer would be x= -6

6 0

3 years ago

Please help need 4 and 5 I will make brainliest , 30 points

trasher [3.6K]

She put 1/5 down which means she owes 4/5

620 * 4/5 = 496

She owes $496.00

98 - 10 = 88

She still owes $88.00

5 0

3 years ago

4. Solve for w:<br> -33 = 6 - 3w

gavmur [86]

Answer:

13 =w

Step-by-step explanation:

-33 = 6 - 3w

Subtract 6 from each side

-33-6 = 6 - 3w-6

-39 = -3w

Divide each side by -3

-39/-3 = -3w/-3

13 =w

3 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