The quotient of a number and 4, minus 2, equals 5/4.<br> Find the number.

Verify that f and g are inverses functions using composition , show your steps

pshichka [43]

Answer:

See explanations below

Step-by-step explanation:

Given the functions

f(x) = 12x - 12

g(x) = x/12 - 1

To show they are inverses, we, must show that f(g(x)) = g(f(x))

f(g(x)) = f(x/12 - 1)

Replace x with x/12 - 1 into f(x)

f(g(x)) =12((x-12)/12) - 11

f(g(x)) = x-1 - 1

f(g(x)) =x - 2

Similarly for g(f(x))

g(f(x)) = g(12x-12)

g(f(x)) =(12x-12)/12 - 1

12(x-1)/12 - 1

x-1 - 1

x - 2

Since f(g(x)) = g(f(x)) = x -2, hence they are inverses of each other

5 0

3 years ago

Witch of the following statements are true

Sliva [168]

It would be B, although it could be different if there’s a picture above

4 0

2 years 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

3 years ago

I need help with the diagram for customary and metric I’m giving 100 points.

sukhopar [10]

Answer:

Customary:

Foot, inch, mile, yard.

Metric:

Centimeter, kilometer, meter.

Both:

Area, height, formula, perimeter.

8 0

3 years ago

Read 2 more answers

How many groups of 1/2 day are in 1 week?

weqwewe [10]

There are 14 groups of days are in 1 week.

5 0

3 years ago

The quotient of a number and 4, minus 2, equals 5/4. Find the number.