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3 years ago

A car going 72.0 km/h (20.0 m/s) slows down to 36.0 km/h (10.0 m/s) in 5.00 seconds. Calculate the acceleration in

Mathematics

1 answer:

Vlada [557]3 years ago

6 0

Answer:

-2 m/s^2

Step-by-step explanation:

Just as you would calculate the velocity when a body moves from point A to point B in a time T, we are going to calculate the acceleration as the difference of the velocities (final - initial) and divide it by the time it took:

Velocity_final = 10 m/s

Velocity_initial = 20 m/s

Time = 5 s

acceleration = (Velocity_final - Velocity_inital)/Time

= (10 m/s - 20 m/s)/5 s

= (-10 m/s) / 5s

= -2 m/s^2

The acceleration in m/s^2 is -2

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Answer:

I believe that Kevin is 6 years old now.

Step-by-step explanation:

I used the guess and check method to figure this out.

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4 years ago

Which statement describes the inverse of m(x) = x^2 – 17x?

DochEvi [55]

Given:

The function is

$m(x)=x^2-17x$

To find:

The inverse of the given function.

Solution:

We have,

$m(x)=x^2-17x$

Substitute m(x)=y.

$y=x^2-17x$

Interchange x and y.

$x=y^2-17y$

Add square of half of coefficient of y , i.e., $\left(\dfrac{-17}{2}\right)^2$ on both sides,

$x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

Taking square root on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}$

Add $\dfrac{17}{2}$ on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y$

Substitute $y=m^{-1}(x)$ .

$m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$

We know that, negative term inside the root is not real number. So,

$x+\left(\dfrac{17}{2}\right)^2\geq 0$

$x\geq -\left(\dfrac{17}{2}\right)^2$

Therefore, the restricted domain is $x\geq -\left(\dfrac{17}{2}\right)^2$ and the inverse function is $m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$ .

Hence, option D is correct.

Note: In all the options square of $\dfrac{17}{2}$ is missing in restricted domain.

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