Leaving Cert Applied Maths Knowledge Checks

KNOWLEDGE CHECKS

We include Knowledge Check quizzes at the end of each of our digital lessons.

Below, you will find a few examples of these quizzes.

/10

\[\,\]Knowledge Check 20

\[\,\]A quick check on your understanding of:

First order differential equations

Second order differential equations

Separation of variables

\[\,\]

A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\).

1 / 10

The equation \(\displaystyle\frac{dy}{dx} = 7\) is a first order differential equation.

True

False

A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\), which is indeed the case for this equation.

Nice! A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\), which is indeed the case for this equation.

A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\), which is indeed the case for this equation.

A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\).

2 / 10

The equation \(\displaystyle\frac{dy}{dx} = \sqrt{2}\sin 5x^3\) is a first order differential equation.

True

False

A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\), which is indeed the case for this equation.

That's right! A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\), which is indeed the case for this equation.

A first order differential equation is any equation of the form \(\displaystyle\frac{dy}{dx}= f(x,y)\), which is indeed the case for this equation.

\(\displaystyle\frac{d^2y}{dx^2}\) is referred to as a second derivative.

3 / 10

If an equation has both a \(\displaystyle\frac{d^2y}{dx^2}\) term and a \(\displaystyle\frac{dy}{dx}\) term, it is considered

Neither a first-order nor second-order differential equation

Only a first-order differential equation

Only a second-order differential equation

Both a first-order and a second-order differential equation

An equation that contains both first and second derivatives is considered a second-order differential equation.

That's the one! An equation that contains both first and second derivatives is considered a second-order differential equation.

An equation that contains both first and second derivatives is considered a second-order differential equation.

\(e^{a+b} = e^ae^b\)

4 / 10

Which of the following equations is equivalent to \(y(x) = e^{x+A}\)?

\(y(x) = Ae^x\)

\(y(x) = Be^x\), where \(B=e^A\)

\(y(x) = e^x +A\)

\(y(x) = e^x + B\), where \(B=e^A\)

\(e^{a+b} = e^ae^b\)

Right answer! \(e^{a+b} = e^ae^b\)

\(e^{a+b} = e^ae^b\)

What is the value of \(y\) when \(x=0\)?

5 / 10

For the function \(y(x)=5x^2+A\), where \(A\) is an unknown constant, which of the following statements correctly describes a graph of this function?

We don't know either the shape or position of this function.

We know the shape of the function but we don't know its position.

We know the position of the function but we don't know its shape.

We know both the shape and position of this function.

As \(y(0)=5(0^2)+A=A\), we do not know the position of this point and therefore the position of the curve. However, we can find the shape of the curve by setting \(A\) to any value, e.g. zero, and plotting points.

Yes! As \(y(0)=5(0^2)+A=A\), we do not know the position of this point and therefore the position of the curve. However, we can find the shape of the curve by setting \(A\) to any value, e.g. zero, and plotting points.

We are free to choose the side(s) to put our constant(s) on.

6 / 10

If \(\displaystyle\frac{dy}{dx}=\frac{10x}{3y}\), which of the following is also true? (Select all that apply.)

\(10x\,dx=3y\, dy\)

\(x\,dx=\displaystyle\frac{3y}{10}\, dy\)

\(10\int x\,dx=3\int y\, dy\)

\(\int x\,dx=\displaystyle\frac{3}{10}\int y\, dy\)

We are free to choose the side(s) to put our constant(s) on.

You got it! We are free to choose the side(s) to put our constant(s) on.

We are free to choose the side(s) to put our constant(s) on.

We can use separation of variables on equations of the form \(\displaystyle\frac{dy}{dx} = f(x)g(y)\).

7 / 10

If we wish to solve the following differential equation using separation of variables, what must be the value of \(a\)?

\begin{align}\displaystyle\frac{dy}{dx} = 5x+y^a\end{align}

-5

We can only use separation of variables on equations of the form \(\displaystyle\frac{dy}{dx} = f(x)g(y)\). If \(a=0\), the equation becomes

\begin{align}\displaystyle\frac{dy}{dx}=5x+1\end{align}

i.e. \(f(x) = 5x+1\) and \(g(y)=1\). For any other value of \(a\), we cannot write the right hand side as \(f(x)g(y)\).

Nice! We can only use separation of variables on equations of the form \(\displaystyle\frac{dy}{dx} = f(x)g(y)\). If \(a=0\), the equation becomes

\begin{align}\displaystyle\frac{dy}{dx}=5x+1\end{align}

i.e. \(f(x) = 5x+1\) and \(g(y)=1\). For any other value of \(a\), we cannot write the right hand side as \(f(x)g(y)\).

We can only use separation of variables on equations of the form \(\displaystyle\frac{dy}{dx} = f(x)g(y)\). If \(a=0\), the equation becomes

\begin{align}\displaystyle\frac{dy}{dx}=5x+1\end{align}

i.e. \(f(x) = 5x+1\) and \(g(y)=1\). For any other value of \(a\), we cannot write the right hand side as \(f(x)g(y)\).

We are given the acceleration rate \(a\) in terms of \(t\) and are asked to find the object's displacement \(s\) at a time \(t\).

8 / 10

"The acceleration rate of an object is described by \(a(t)=5t\). Find an equation for the object's displacement at time \(t\)."

\[\,\]

Which of the following equations should we solve?

\(a=\displaystyle\frac{dv}{dt}\)

\(a=v\displaystyle\frac{dv}{ds}\)

\(a=\displaystyle\frac{d^2s}{dt^2}\)

\(v=\displaystyle\frac{ds}{dt}\)

We are given the acceleration rate \(a\) in terms of \(t\) and are asked to find the object's displacement \(s\) at a time \(t\). Therefore, we should solve the differential equation that contains \((a,s,t)\).

Score! We are given the acceleration rate \(a\) in terms of \(t\) and are asked to find the object's displacement \(s\) at a time \(t\). Therefore, we should solve the differential equation that contains \((a,s,t)\).

We are given the acceleration rate \(a\) in terms of \(t\) and are asked to find the object's speed \(v\) at a time \(t\).

9 / 10

"The acceleration rate of an object is described by \(a(t)=5t\). Find an equation for the object's speed at time \(t\)."

\[\,\]

Which of the following equations should we solve?

\(a=\displaystyle\frac{dv}{dt}\)

\(a=v\displaystyle\frac{dv}{ds}\)

\(a=\displaystyle\frac{d^2s}{dt^2}\)

\(v=\displaystyle\frac{ds}{dt}\)

We are given the acceleration rate \(a\) in terms of \(t\) and are asked to find the object's speed \(v\) at a time \(t\). Therefore, we should solve the differential equation that contains \((a,v,t)\).

Ding ding! We are given the acceleration rate \(a\) in terms of \(t\) and are asked to find the object's speed \(v\) at a time \(t\). Therefore, we should solve the differential equation that contains \((a,v,t)\).

We are given the acceleration rate \(a\) in terms of \(s\) and are asked to find the object's speed \(v\) at a certain displacement \(s\).

10 / 10

"The acceleration rate of an object is described by \(a(t)=5s\). Find an equation for the object's speed when it has been displaced by an amount \(s\) from its original location."

\[\,\]

Which of the following equations should we solve?

\(a=\displaystyle\frac{dv}{dt}\)

\(a=v\displaystyle\frac{dv}{ds}\)

\(a=\displaystyle\frac{d^2s}{dt^2}\)

\(v=\displaystyle\frac{ds}{dt}\)

We are given the acceleration rate \(a\) in terms of \(s\) and are asked to find the object's speed \(v\) at a certain displacement \(s\). Therefore, we should solve the differential equation that contains \((a,v,s)\).

Excellent! We are given the acceleration rate \(a\) in terms of \(s\) and are asked to find the object's speed \(v\) at a certain displacement \(s\). Therefore, we should solve the differential equation that contains \((a,v,s)\).

Your score is

/10

\[\,\]Knowledge Check 40

\[\,\]A quick check on your understanding of:

Arithmetic series

Geometric series

\[\,\]

\(S_n = \dfrac{n}{2}[2a+(n-1)d]\)

1 / 10

The general term for a linear sequence is shown below

\begin{align}T_n = n+3\end{align}

What is the sum of the first six terms of this sequence?

As \(a=T_1=1+3=4\), \(d=1\) and \(n=6\), we obtain

\begin{align}S_6 &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{6}{2}[2(4)+(6-1)(1)]\\&=39\end{align}

Right! As \(a=T_1=1+3=4\), \(d=1\) and \(n=6\), we obtain

\begin{align}S_6 &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{6}{2}[2(4)+(6-1)(1)]\\&=39\end{align}

As \(a=T_1=1+3=4\), \(d=1\) and \(n=6\), we obtain

\begin{align}S_6 &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{6}{2}[2(4)+(6-1)(1)]\\&=39\end{align}

\(S_n = \dfrac{n}{2}[2a+(n-1)d]\)

2 / 10

The general term for a linear sequence is shown below

\begin{align}T_n = n+3\end{align}

What is the sum of the first twenty terms of this sequence?

As \(a=T_1=1+3=4\), \(d=1\) and \(n=20\), we obtain

\begin{align}S_{20} &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{20}{2}[2(4)+(20-1)(1)]\\&=270\end{align}

Well done! As \(a=T_1=1+3=4\), \(d=1\) and \(n=20\), we obtain

\begin{align}S_{20} &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{20}{2}[2(4)+(20-1)(1)]\\&=270\end{align}

As \(a=T_1=1+3=4\), \(d=1\) and \(n=20\), we obtain

\begin{align}S_{20} &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{20}{2}[2(4)+(20-1)(1)]\\&=270\end{align}

\(S_n = \dfrac{n}{2}[2a+(n-1)d]\)

3 / 10

The general term for a linear sequence is shown below

\begin{align}T_n = 5n-2\end{align}

What is the sum of the first six terms of this sequence?

As \(a=T_1=5(1)-2=3\), \(d=5\) and \(n=6\), we obtain

\begin{align}S_6 &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{6}{2}[2(3)+(6-1)(5)]\\&=93\end{align}

Got it! As \(a=T_1=5(1)-2=3\), \(d=5\) and \(n=6\), we obtain

\begin{align}S_6 &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{6}{2}[2(3)+(6-1)(5)]\\&=93\end{align}

As \(a=T_1=5(1)-2=3\), \(d=5\) and \(n=6\), we obtain

\begin{align}S_6 &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{6}{2}[2(3)+(6-1)(5)]\\&=93\end{align}

\(S_n = \dfrac{n}{2}[2a+(n-1)d]\)

4 / 10

The general term for a linear sequence is shown below

\begin{align}T_n = 5n-2\end{align}

What is the sum of the first twenty terms of this sequence?

As \(a=T_1=5(1)-2=3\), \(d=5\) and \(n=20\), we obtain

\begin{align}S_{20} &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{20}{2}[2(3)+(20-1)(5)]\\&=1{,}010\end{align}

Superb! As \(a=T_1=5(1)-2=3\), \(d=5\) and \(n=20\), we obtain

\begin{align}S_{20} &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{20}{2}[2(3)+(20-1)(5)]\\&=1{,}010\end{align}

As \(a=T_1=5(1)-2=3\), \(d=5\) and \(n=20\), we obtain

\begin{align}S_{20} &= \dfrac{n}{2}[2a+(n-1)d]\\&=\dfrac{20}{2}[2(3)+(20-1)(5)]\\&=1{,}010\end{align}

This is an arithmetic (linear) sequence.

5 / 10

If the first difference between successive terms is \(0.25\), that sequence will converge.

True

False

If the difference between successive terms is a constant, that sequence is arithmetic (linear). However, there is no corresponding arithmetic series that converges. Instead, it is only (some) geometric series which converge.

Well done! If the difference between successive terms is a constant, that sequence is arithmetic (linear). However, there is no corresponding arithmetic series that converges. Instead, it is only (some) geometric series which converge.

\(S_n = \dfrac{a(1-r^n)}{1-r}\)

6 / 10

The general term for a geometric sequence is shown below

\begin{align}T_n = 2(3^{n-1})\end{align}

What is the sum of the first six terms of this sequence?

As \(a=2\), \(r=3\) and \(n=10\), we obtain

\begin{align}S_{10} &= \dfrac{a(1-r^n)}{1-r}\\&=\dfrac{2(1-3^{6})}{1-3}\\&=728\end{align}

Excellent!

As \(a=2\), \(r=3\) and \(n=10\), we obtain

\begin{align}S_{10} &= \dfrac{a(1-r^n)}{1-r}\\&=\dfrac{2(1-3^{6})}{1-3}\\&=728\end{align}

As \(a=2\), \(r=3\) and \(n=10\), we obtain

\begin{align}S_{10} &= \dfrac{a(1-r^n)}{1-r}\\&=\dfrac{2(1-3^{6})}{1-3}\\&=728\end{align}

The common ratio for this sequence is \(3\).

7 / 10

The general term for a geometric sequence is shown below

\begin{align}T_n = 2(3^{n-1})\end{align}

What does the corresponding series converge to?

\(750\)

\(875\)

\(916\)

It does not converge.

As the common ratio for this sequence is \(3\), it will not converge (as the common ration must be between \(0\) and \(1\)).

Nice! As the common ratio for this sequence is \(3\), it will not converge (as the common ration must be between \(0\) and \(1\)).

As the common ratio for this sequence is \(3\), it will not converge (as the common ration must be between \(0\) and \(1\)).

\(S_n = \dfrac{a(1-r^n)}{1-r}\)

8 / 10

The general term for a geometric sequence is shown below

\begin{align}T_n = 10(0.2^{n-1})\end{align}

What is the sum of the first four terms of this sequence?

\(12.12\)

\(12.24\)

\(12.36\)

\(12.48\)

As \(a=10\), \(r=0.2\) and \(n=4\), we obtain

\begin{align}S_4 &= \dfrac{a(1-r^n)}{1-r}\\&=\dfrac{10(1-0.2^{4})}{1-0.2}\\&=12.48\end{align}

Ding! As \(a=10\), \(r=0.2\) and \(n=4\), we obtain

\begin{align}S_4 &= \dfrac{a(1-r^n)}{1-r}\\&=\dfrac{10(1-0.2^{4})}{1-0.2}\\&=12.48\end{align}

As \(a=10\), \(r=0.2\) and \(n=4\), we obtain

\begin{align}S_4 &= \dfrac{a(1-r^n)}{1-r}\\&=\dfrac{10(1-0.2^{4})}{1-0.2}\\&=12.48\end{align}

The common ratio for this sequence is \(0.2\).

9 / 10

The general term for a geometric sequence is shown below

\begin{align}T_n = 10(0.2^{n-1})\end{align}

What does the corresponding series converge to?

\(10\)

\(12.5\)

\(15.25\)

\(16.4\)

As the common ratio for this sequence is \(0.2\), it will converge.

As \(a=10\) and \(r=0.2\), this series will converge to the following value:

\begin{align}S_\infty &= \frac{a}{1-r}\\&=\frac{10}{1-0.2}\\&=12.5\end{align}

Bravo! As the common ratio for this sequence is \(0.2\), it will converge.

As \(a=10\) and \(r=0.2\), this series will converge to the following value:

\begin{align}S_\infty &= \frac{a}{1-r}\\&=\frac{10}{1-0.2}\\&=12.5\end{align}

As the common ratio for this sequence is \(0.2\), it will converge.

As \(a=10\) and \(r=0.2\), this series will converge to the following value:

\begin{align}S_\infty &= \frac{a}{1-r}\\&=\frac{10}{1-0.2}\\&=12.5\end{align}

This sequence has a common ration of \(2\).

10 / 10

Which of the following statements are true about the following sequence? (Select all the apply.)

\begin{align}1,2,4,8,16,...\end{align}

It is an arithmetic sequence.

It is a geometric sequence.

Its corresponding series converges.

Its corresponding series does not converge.

This sequence has a common ratio of \(2\) and is therefore a geometric sequence. As this ratio is not between \(0\) and \(1\), its corresponding series does not converge.

Right answer! This sequence has a common ratio of \(2\) and is therefore a geometric sequence. As this ratio is not between \(0\) and \(1\), its corresponding series does not converge.

This sequence has a common ratio of \(2\) and is therefore a geometric sequence. As this ratio is not between \(0\) and \(1\), its corresponding series does not converge.

Your score is

/10

\[\,\]Knowledge Check 50

\[\,\]A quick check on your understanding of:

Routing problems

Allocation of resources

Equipment replacement and maintenance

Stock control

\[\,\]

The states are the number of items in stock.

1 / 10

In a stock control question, if a state has a \(0\) label, what does that imply?

There is currently no stock.

We currently have no orders.

We cannot fulfil an order.

We are currently breaking even.

The states are the number of items in stock.

Right! The states are the number of items in stock.

The states are the number of items in stock.

We cross out a state if there are no possible actions from that state.

2 / 10

In a stock control question, we cross out a state if:

We have no stock.

We currently have no orders.

We cannot fulfil an order.

We are currently breaking even.

If there is not way to fulfil an upcoming order if we are in this state, we cross out that state as there are no possible actions from that state.

Very good! If there is not way to fulfil an upcoming order if we are in this state, we cross out that state as there are no possible actions from that state.

If there is not way to fulfil an upcoming order if we are in this state, we cross out that state as there are no possible actions from that state.

Can we have surplus stock?

3 / 10

In a stock control question, the largest value that a state can have is the largest value in our order book.

True

False

We can typically also store surplus stock (at a cost).

Ding! We can typically also store surplus stock (at a cost).

We can typically also store surplus stock (at a cost).

The weights of each action is the total amount of money made at each event.

4 / 10

For a routing problem, we minimise the value as we want the travelling costs to be as low as possible.

True

False

We are instead trying to maximise the total profit made across all events, taking into account the travelling costs.

Good! We are instead trying to maximise the total profit made across all events, taking into account the travelling costs.

We are instead trying to maximise the total profit made across all events, taking into account the travelling costs.

What events are the most profitable each day?

5 / 10

A promoter is setting up a food tent at some of the following events.

\[\,\]

Only one event can be attended per day. The expected profits at each event are shown below.

\[\,\]

If travelling costs are negligible, what events should the promoter attend to maximise profits?

\(A\), \(E\) and \(G\)

\(B\), \(C\) and \(E\)

\(A\), \(C\) and \(E\)

\(C\), \(E\) and \(G\)

Event \(A\) is the only event on day one, event \(C\) is the most profitable on day two and event \(E\) is the most profitable on day three.

Great! Event \(A\) is the only event on day one, event \(C\) is the most profitable on day two and event \(E\) is the most profitable on day three.

Event \(A\) is the only event on day one, event \(C\) is the most profitable on day two and event \(E\) is the most profitable on day three.

The optimal events to attend are \(A\), \(C\) and \(E\).

6 / 10

A promoter is setting up a food tent at some of the following events.

\[\,\]

Only one event can be attended per day. The expected profits at each event are shown below.

\[\,\]

If travelling costs are negligible, what is the maximum amount of profit that the promoter can make at these events?

€\(210\)

€\(230\)

€\(270\)

€\(300\)

The optimal events to attend are \(A\), \(C\) and \(E\), each of which turn a profit of €\(30\), €\(80\) and €\(120\) respectively.

Bingo! The optimal events to attend are \(A\), \(C\) and \(E\), each of which turn a profit of €\(30\), €\(80\) and €\(120\) respectively.

The optimal events to attend are \(A\), \(C\) and \(E\), each of which turn a profit of €\(30\), €\(80\) and €\(120\) respectively.

A sink state exists only if it is the destination for all actions of that stage.

7 / 10

Rebecca has purchased a new bicycle at a cost of €\(500\).

Her bicycle will depreciate in value over time. She keeps any bicycle for a maximum for \(4\) years.

The maintenance cost to keep the bicycle roadworthy each year depends on the age of the bicycle.

Any time Rebecca sells her bicycle, she will repurchase the same bicycle brand new.

After \(10\) years, the corresponding network will have a sink node only if:

She bought a bicycle at the start of that year and sold it at the end of that year.

She bought a bicycle at the end of that year but did not sell it the next year.

She sold a bicycle at the end of that year and bought a new one at the start of the next year.

She sold a bicycle at the end of that year and did not buy a new one at the start of the next year.

In this case, all actions for this stage will have same destination, i.e. a sink state.

Bravo! In this case, all actions for this stage will have same destination, i.e. a sink state.

In this case, all actions for this stage will have same destination, i.e. a sink state.

Rebecca is expected to make a loss overall.

8 / 10

Rebecca has purchased a new bicycle at a cost of €\(500\).

Her bicycle will depreciate in value over time. She keeps any bicycle for a maximum for \(4\) years.

The maintenance cost to keep the bicycle roadworthy each year depends on the age of the bicycle.

Any time Rebecca sells her bicycle, she will repurchase the same bicycle brand new.

After \(10\) years, Rebecca sells her bicycle as she now prefers jogging.

If we use dynamic programming to determine when and how often Rebecca should sell her bicycle to minimise the amount of money she spends within these 10 years, which of the following should we do?

Find the minimum value of that network, which will be negative.

Find the minimum value of that network, which will be positive.

Find the maximum value of that network, which will be negative.

Find the maximum value of that network, which will be positive.

We should try to maximise her profit. However, she is expected to make a loss overall by selling used bicycles and buying brand new ones.

Score! We should try to maximise her profit. However, she is expected to make a loss overall by selling used bicycles and buying brand new ones.

We should try to maximise her profit. However, she is expected to make a loss overall by selling used bicycles and buying brand new ones.

James always uses at least \(100\) kilograms to make bread each week.

9 / 10

James harvests \(500\) kilograms of wheat on his farm each week.

Rather than directly selling it as wheat, he uses all of it to instead produce three products (bread, pasta and crackers) which he then sells.

The amount of profit (in euro) that he makes per \(100\) kilograms used for each product is as shown.

\[\,\]

James always uses at least \(100\) kilograms to make bread each week.

Unfortunately, due to a drought, James instead only harvested \(200\) kilograms of wheat this week. What should he make to maximise his profit?

Bread only

Bread and pasta

Bread and crackers

Pasta and crackers

James always uses at least \(100\) kilograms to make bread each week. As we have \(100\) kilograms of wheat left over, we should use it to make crackers as it is the most profitable according to the first column.

Excellent! James always uses at least \(100\) kilograms to make bread each week. As we have \(100\) kilograms of wheat left over, we should use it to make crackers as it is the most profitable according to the first column.

What product do we place in the first stage of the Guided Example?

10 / 10

James harvests \(500\) kilograms of wheat on his farm each week.

Rather than directly selling it as wheat, he uses all of it to instead produce three products (bread, pasta and crackers) which he then sells.

The amount of profit (in euro) that he makes per \(100\) kilograms used for each product is as shown.

\[\,\]

James always uses at least \(100\) kilograms to make bread each week.

If we wish to use dynamic programming to determine how James should use his supply of wheat to maximise his profit, which product should we use in the first stage?

Bread (as it minimises the number of actions in later stages)

Pasta (as it is the most profitable in the final column)

Crackers (as it is the most profitable in the first column)

By using bread as the first stage, we only have at most \(400\) kilograms remaining to distribute in the following stages.

Got it! By using bread as the first stage, we only have at most \(400\) kilograms remaining to distribute in the following stages.

By using bread as the first stage, we only have at most \(400\) kilograms remaining to distribute in the following stages.

Your score is

KNOWLEDGE CHECKS

We include Knowledge Check quizzes at the end of each of our digital lessons.

Below, you will find a few examples of these quizzes.

Free Resources

Links

Contact

Our Websites