All categories

3 years ago

F(x)= 10x-5What is the value of f-1(-4) ?

Physics

2 answers:

Afina-wow [57]3 years ago

6 0

Answer:

$f^{-1}(-4) = \frac{1}{10}$

Explanation:

Firstly finding $f^{-1}(x)$

So,

$f(x) = 10x-5$

Substitute $y = f(x)$

$y = 10x-5$

Exchange the values of x and y

$x = 10y-5$

Solving for y

$x = 10y-5$

Adding 5 to both sides

$10y = x+5$

Dividing both sides by 10

$y = \frac{x+5}{10}$

Replace $y = f^{-1}(x)$

$f^{-1}(x) = \frac{x+5}{10}$

For x = -4

$f^{-1}(-4) = \frac{-4+5}{10}$

$f^{-1}(-4) = \frac{1}{10}$

Eva8 [605]3 years ago

4 0

Answer:

$\frac{1}{10}$

Explanation:

f(x) = y (output)

y = 10x - 5

Switch variables.

Solve for y.

x = 10y - 5

x + 5 = 10y

x/10 + 1/2 = y

$f^{-1}(x)$ = 1/10x + 1/2

Put x as -4.

1/10(-4) + 1/2

-4/10 + 1/2

-4/10 + 5/10

= 1/10

You might be interested in

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0

3 years ago

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball

Fudgin [204]

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

$V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\ mm^3$

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

$n = 520000\times 0.523\ grains\\\\n = 271960\ grains$

8 0

3 years ago

Block A of 5 kg with a speed of 3 m/s collides with block B of 10 kg that is stationary. After the collision, block B travels wi

Lemur [1.5K]

Answer:

-1m/s

Explanation:

We can calculate the speed of block A after collision

According to collision theory:

MaVa+MbVb = MaVa+MbVb (after collision)

Substitute the given values

5(3)+10(0) = 5Va+10(2)

15+0 = 5Va + 20

5Va = 15-20

5Va = -5

Va = -5/5

Va = -1m/s

Hence the velocity of ball A after collision is -1m/s

Note that the velocity of block B is zero before collision since it is stationary

6 0

3 years ago

how much force is needed to cause a 15 kilogram bike to accelerate at a rate of 10 meters per second?

egoroff_w [7]

F = m*a, mass times acceleration.

F = 15*10 = 150 N

8 0

3 years ago

Read 2 more answers

A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?

Olenka [21]

Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
$F = k*x$

Solving:
$F = k*x$
$F = 50*0.15$
$\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark$

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
$E = \frac{k*x^2}{2}$

Solving:(Energy associated with this stretching)
$E = \frac{k*x^2}{2}$
$E = \frac{50*0.15^2}{2}$
$E = \frac{50*0.0225}{2}$
$E = \frac{1.125}{2}$
$\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark$

7 0

3 years ago

F(x)= 10x-5What is the value of f-1(-4) ?​

F(x)= 10x-5What is the value of f-1(-4) ?