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2 years ago

15

3x + 5=. X equals six

1 answer:

horsena [70]2 years ago

8 0

Answer:

23

Step-by-step explanation:

3 (6) + 5 =

You would have to multiply first

So,

<h2>3 x 6 = <u><em>18</em></u></h2>

Then add

<h2><u><em>18 </em></u>+ 5 = 23</h2>

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Tres hombres forman una sociedad y se ponen de acuerdo para repartir las ganancias en partes iguales x invierte $3500 y Y invier

Arlecino [84]

For this case, the total amount invested is:
investment = 3500 + 1000 = 4500
We must take out the percentage of what X invested.
X investment = (3500/4500) * 100 = 77.78%.
Therefore, X gets 77.78% of the profits:
X profit = 0.7778 * 3000 = 2333.33 $
answer
if the profits are distributed in proportion to the amount invested X receives 2333.33 $

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3 years ago

The coordinate grid shows the plot of four equations. a coordinate plane is shown with four lines graphed. line a crosses the x

hammer [34]

Its A & B

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2 years ago

Read 2 more answers

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Delvig [45]

Answer:I believe the answer is A

Step-by-step explanation:

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1 year 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

2 years ago

11. You plan to bake "apple crisp" for a dinner party you are hosting. Your recipe serves four

Ludmilka [50]

Answer:

2.5 teaspoon.

Step-by-step explanation:

4 serve recipe need cinnamon = 1 tsp

1 serve recipe need cinnamon = 1/4 tsp

10 serve recipe need cinnamon = 1/4 * 10

= 10/4

= 2.5 tsp

7 0

3 years ago