We can see from the graph, as the value of x approaches 2 from the left hand side i.e. the negative side, the value of the function approaches 4.
As the value of x approaches 2 from the positive side i.e. the right hand side, the value of function approaches -4
Since the Right hand limit and the Left hand limit as x approaches 2 are not the same, the limit does not exist at x = 2
Answer:
<em>It results -14 in either way</em>
Step-by-step explanation:
<u>Velocity As A Rate Of Change
</u>
The velocity of an object can be computed as the rate of change of its displacement (or position taken as a vector) over time. If we compute it as a derivative, it's called instantaneous velocity, and if computed as the slope of the function (difference quotient) at a certain point it's the average velocity
The position of the object as a function of time is
Computing the derivative
We can see it's a constant value. If we use the slope or rate of change:
Now let's fix two values for time
and compute the corresponding positions, by using the given function
Now we compute the average velocity
We get the very same result in both ways to compute v. It happens because the position is related with time as a linear function, it's called a constant velocity motion.
Answer:
7
Step-by-step explanation:
16-9=7
parralel lines
I kNoW I SPelLed dat WrOnG
Answer:
<h2>SU = 3.5</h2>
Step-by-step explanation:
We have the triangle 30° - 60° - 90°.
Right triangles with 30-60-90 angles will have their ratio of the sides as
1 : √3 : 2 (look at the picture).
We have:
SR : SU : RU = 1 : √3 : 2
Therefore RU = 2SR and SU = SR√3.
We have RU = 4. Substitute:
2SR = 4 <em>divide both sides by 2</em>
SR = 2
SU = 2√3 ≈ 2(1.73) = 3.5
The new pressure is 2 Kpa
<em><u>Solution:</u></em>
At constant temperature, the pressure of a sample of gas is inversely proportional to its volume
Therefore,
Let "p" be the pressure of sample gas
Let "v" be the volume
Then we get,
Where, "k" is the constant of proportionality
<em><u>I have some oxygen in a 2.28 liter container with a pressure of 5 kPa</u></em>
Substitute v = 2.28 and p = 5 in eqn 1
<em><u>I move all of it to a 5.7 liter container at the same temperature</u></em>
Substitute k = 11.4 and v = 5.7 in eqn 1
Thus new pressure is 2 kPa