All categories

2 years ago

Find the value of 17 33 rounded to the nearest ten-thousandth.

Mathematics

1 answer:

alexira [117]2 years ago

7 0

Answer:

0.5152

Step-by-step explanation:

You might be interested in

What is the answer if you put a link you are getting reported

Alinara [238K]

Answer:

its more than 3/4 :)

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

I'm learning about Equivalent systems of equations, please help!

kobusy [5.1K]

OK to solve this, we have to solve each system presented through elimination or substitution and find which one is equivalent to that of the teacher's!

First let's solve for the teacher's:
-2x+5y=10
-3x+9y=6
Solve by substitution (I think elimination might be easier to do for this one, but I don't really remember 100% sorry!)
Isolate the x (or y) variable in the first equation
-2x+5y=10
-2x=10-5y
$x= \frac{10-5y}{-2}$
Substitute x into the next equation and solve for y
-3(10-5y/2)+9y=6
3*10-5y/2+9y=6
(multiply both sides by 2)
3(10-5y)+18y=12
30-15y+18y=12
30+3y=12
3y=-18
y=-6

Substitute in x
x= -10-5(-6)/2
x=-20

TEACHER'S ANSWER (-20,-6)

GOKU
x-3y=-2
-2x+5y=-7
Do the same as above
Solve for x
x-3y=-2
x=3y-2
Plug in
-2(3y-2)+5y=-7
4-6y+5y=-7
4-y=-7
-y=-11
y=11

x=(3(11)-2)
x=31
GOKU'S ANSWER (31, 11)

SELINA:
-5x+14y=16
-3x+9y=12
One last time!! :)

-5x+14y=16
-5x=16-14y
x=(16-14y)/-5

-3(-(16-14y/5)+9y=12
3*16-14y/5+9y=12
3*16-14y+45y=60
48-42y+45y=60
48+3y=60
3y=12
y=4

x=-(16-14(4))/5
x=8
SELINA'S ANSWER
(8,4)

So neither Goku or Selina got the same answer as the teacher

3 0

2 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

2 years ago

Suppose there are 100 trout in a lake and the yearly growth factor for the populations is 1.5. Write the exponential equation fo

RSB [31]

T=100(1.5^t)
Where T is number of trout and t is time in years applied as an exponent to the 1.5 factor

5 0

2 years ago

Tell which set or sets of numbers below belongs to: natural numbers, whole numbers, integers, rational numbers, irrational numbe

Misha Larkins [42]

1/3 belongs to the rational set and to the real set.

<h3>To which sets does the number below belong?</h3>

Here we have the number 1/3.

First, remember that we define rational numbers as these numbers that can be written as a quotient between two integers.

Here 1 is an integer and 3 is an integer, then 1/3 is a rational number.

Also, the combination between the rational set and the irrational set is the set of the real numbers, then 1/3 is also a real number.

Then, concluding:

1/3 belongs to the rational set and to the real set.

If you want to learn more about rational numbers:

brainly.com/question/12088221

#SPJ1

6 0

1 year ago