Welcome to our community

Be a part of something great, join today!

Yuri's questions at Yahoo! Answers involving algebra

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here are the questions:

Maths problem please help me?

1.) A liquid flows from container A to B. The volumes, in cm^3 of the liquid in the containers A and B, t seconds after the start of the experiment are

A(t)=-t^3+at^2+p
B(t)=bt(t-2)^2+2t^2+ct+q

At the start of the experiment container A has 300cm^3 and B is empty. Find the values of p and q.

Explain why 300-A(t)=B(t) is in identity for all non-negative values of t up to the instant when container A is empty.( I don't know what the question means)

Use the identity above to find value of a, b and c

2. Factorize (2n+1)^3-(2n-1)^3 completely and show that the difference between the cubes of two consecutive positive odd numbers can never be divisible by 4.
I have posted a link to this topic so the OP can see my work.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Yuri,

1.) We are given the functions:

\(\displaystyle A(t)=-t^3+at^2+p\)

\(\displaystyle B(t)=bt(t-2)^2+2t^2+ct+q\)

Now, we are also given:

\(\displaystyle A(0)=300=-(0)^3+a(0)^2+p=p\,\therefore\,p=300\)

\(\displaystyle B(0)=0=b(0)(0-2)^2+2(0)^2+c(0)+q=q\,\therefore\,q=0\)

Thus, we have:

\(\displaystyle A(t)=-t^3+at^2+300\)

\(\displaystyle B(t)=bt(t-2)^2+2t^2+ct\)

Because the amount of liquid present in A plus the amount present in B has to remain constant (assuming so spills, leaks, evaporation, etc.), we know:

\(\displaystyle A(t)+B(t)=300\)

Which we can arrange as:

\(\displaystyle 300-A(t)=B(t)\)

Using the definitions of the functions, we may write:

\(\displaystyle 300-\left(-t^3+at^2+300 \right)=bt(t-2)^2+2t^2+ct\)

Distributing, expanding and collecting like terms, we have:

\(\displaystyle t^3-at^2=bt^3+(2-4b)t^2+(4b+c)t\)

Equating coefficients, we find:

\(\displaystyle b=1\)

\(\displaystyle 4b-2=a\,\therefore\,a=2\)

\(\displaystyle 4b+c=0\,\therefore\,c=-4\)

2.) Using the difference of cubes formula, we may write:

\(\displaystyle (2n+1)^3-(2n-1)^3=((2n+1)-(2n-1))\left((2n+1)^2+(2n+1)(2n-1)+(2n-1)^2 \right)=2\left(12n^2+1 \right)\)

The second factor was obtained by noting:

\(\displaystyle (2n+1)^2=4n^2+4n+1\)

\(\displaystyle (2n+1)(2n-1)=4n^2-1\)

\(\displaystyle (2n-1)^2=4n^2-4n+1\)

Adding, we find:

\(\displaystyle (2n+1)^2+(2n+1)(2n-1)+(2n-1)^2=12n^2+1\)

Since \(\displaystyle 12n^2+1\) is odd for any $n\in\mathbb{N}$, we find then that the difference between the cubes of two consecutive positive odd numbers can never be divisible by 4.

If we let \(\displaystyle m=6n^2\) then \(\displaystyle 12n^2+1=2m+1\), which is obviously odd.