the final speed in m/s of the 10.0 kg is 2.53 m/s .
<u>Step-by-step explanation:</u>
Here we have , A 10.0 kg and a 2.0 kg cart approach each other on a horizontal friction less air track. Their total kinetic energy before collision is 96 ). Assume their collision is elastic. We need to find What is the final speed in m/s of the 10.0 kg mass if that of the 2.0 kg mass is 8.0 m/s . Let's find out:
We know that in an elastic collision :
⇒ Total kinetic energy before collision = Total kinetic energy after collision
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Therefore , the final speed in m/s of the 10.0 kg is 2.53 m/s .
Answer:
m equals √16, n= 3 1/2
Step-by-step explanation:
the one from the top right
(r-c)(3) means that you will simply plug in the value 3 as x in the functions r(x) and c(x) and take the difference of both.
Solving for the value of the functions we'll get:
Since the value is positive, we can say that the store will profit in its third month. Knowing that r(x) and c(x) is measured in hundreds of dollars, we can also say that the profit will be $3400.
ANSWER: C. The new store will have a profit of $3400 after its third month in business.
i) The given function is
The domain is
ii) For vertical asymptotes, we simplify the function to get;
The vertical asymptote occurs at
iii) The roots are the x-intercepts of the reduced fraction.
Equate the numerator of the reduced fraction to zero.
iv) To find the y-intercept, we substitute 
into the reduced fraction.
v) The horizontal asymptote is given by;
The horizontal asymptote is 
.
vi) The function has a hole at 
.
Thus at 
.
This is the factor common to both the numerator and the denominator.
vii) The function is a proper rational function.
Proper rational functions do not have oblique asymptotes.
