|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Further Mathematics||GCSE||£18 /hr|
|Further Mathematics||IB||£20 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|HL Further Mathematics||Baccalaureate||7|
|SL Ancient Greek||Baccalaureate||6|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
First, we are going to give our decimal a name, let's call it x. We notice that x repeats after two digits, so how about we look at 100x.
100x = 54.54545454...
But, the decimal part of this number is what we called x, so let's subtract it.
100x - x = 54.5454545454... - 0.54545454
Working out both sides, we get
99x = 54
So, x = 54/99. However, we are not finished as the question said "in its simplest form". Either notice that 99 and 54 are both divisible by 9, or that they're divisible by 3 twice and we get the final answer.
x = 6/11see more
For this question, we want to start by finding the possible values of a from the first equation and then using that to give us information about b from the second equation.
Now, the condition that a^2 < 4 actually gives us two pieces of information (a way to remember this is that for inequalities, squared gives us two things). We have that a < 2 but also that a > -2 since for example, (-3)^2 = 9 which is larger than 4. Therefore, we have -2 < a < 2.
Rearranging the second equation for b, we get b = (8-a)/2. Therefore, putting in our values for a, we get that the maximum value of b is (8-(-2))/2 = 10/2 = 5 and the minimum value of b is (8-2)/2 = 6/2 = 3. Since the inequality for a does not include 2 and -2, we do not include 3 and 5 for b and so we get the answer 3 < b < 5.see more
Here, we must remember the definition of a bijection. To be bijective, a function must be both injective and surjective. For a function to be injective, we must have that f(a) = f(b) implies that a = b. For a function to be surjective, we must have that for all elements of the range (right hand side), there is an element of the domain (left hand side) that is sent to it by the function.
So, let's start by showing injectivity. We take two pairs (p,q) and (r,s) such that f(p,q) = f(r,s).
We have 2p + q = 2r + s (1) and p + q = r + s (2). I have labelled them (1) and (2) for convenience. Now we look at (1) - (2), and get p = r. Applying this to equation (2), q = s. Therefore, (p,q) = (r,s), our function is injective.
Next, we must consider surjectivity. Let us take a point (a,b) in Z x Z. If 2x + y = a and x + y = b, then as before we can find that x = a - b and y = 2b - a. Therefore, for any a and b in Z x Z, we have found values of x and y that map to it. Since a and b are integers, x and y must also be integers. Therefore our function is surjective as required.
Combining the two results, we have that the function is bijective.see more