For some of my math i just cant figure out what i have to do just to start
Any help or review to show me how to do some would be great?
here is a link to the questions
http://www.math.unb.ca/ready/paper/
For some of my math i just cant figure out what i have to do just to start
Any help or review to show me how to do some would be great?
here is a link to the questions
http://www.math.unb.ca/ready/paper/
There is a lot here, and it would be a mistake for us to do them all for you.Originally Posted by asdfghjkl
But I see no reason for not showing you how to do a selection of these.
1a. Simplify: $\displaystyle \frac{x^3-9x}{x^2-7x+12}$.
First if $\displaystyle x$ is a common factor of the top or bottom of a
rational expression like this, factor it out:
$\displaystyle \frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $.
Now look for terms which are the difference of two squares as we know
that $\displaystyle a^2+b^2=(a+b)(a-b)$, here we have such a term, so:
$\displaystyle
\frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $
$\displaystyle
=\frac{x(x+3)(x-3)}{x^2-7x+12}
$
Now we check if we can cancel any terms, in this case we check if:
$\displaystyle x^2-7x+12$,
has one or more of $\displaystyle x$, $\displaystyle x+3$ or $\displaystyle x-3$
as factors. The first obviously is not a factor, if the second was a factor
then substituting $\displaystyle x=-3$ into the expression would simplify to 0,
it does not so $\displaystyle x+3$ id not a factor of $\displaystyle x^2-7x+12$.
Now if the last candidate is a factor substituting $\displaystyle x=3$ into the
expressions would simplify to zero, which in this case it does, and by trial
and error we find:
$\displaystyle x^2-7x+12=(x-3)(x-4)$, so now we have:
$\displaystyle
\frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $
$\displaystyle.
=\frac{x(x+3)(x-3)}{x^2-7x+12}
$$\displaystyle
=\frac{x(x+3)(x-3)}{(x-3)(x-4)}
$
Finally cancel common terms to get:
$\displaystyle
\frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} $
$\displaystyle.
=\frac{x(x+3)(x-3)}{x^2-7x+12}
$$\displaystyle
=\frac{x(x+3)(x-3)}{(x-3)(x-4)}
$$\displaystyle
=\frac{x(x+3)}{x-4}
$
RonL
2a. Rationalise the denominator of:Originally Posted by asdfghjkl
$\displaystyle
\frac{2}{\sqrt{3}+\sqrt{2}}\ \ \ \ ...(1)
$
Now remember what we said earlier about the difference od two squares:
$\displaystyle
a^2-b^2=(a+b)(a-b)
$
If we put $\displaystyle a=\sqrt{3}$ and $\displaystyle b=\sqrt{2}$, we have:
$\displaystyle
3-2=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})
$
So if we now multiply the top and bottom of $\displaystyle (1)$ by $\displaystyle \sqrt{3}-\sqrt{2}$ we don't change its value so:
$\displaystyle
\frac{2}{\sqrt{3}+\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} $$\displaystyle
=\frac{2\sqrt{3}-2\sqrt{2}}{3-2}=2\sqrt{3}-2\sqrt{2}
$
RonL
4a. Solve for x (without using a claculator)
$\displaystyle
5^{x+1}=25
$
$\displaystyle 25=5\times 5=5^2$, so:
$\displaystyle
5^{x+1}=5^2
$.
Therefore equating exponents $\displaystyle x+1=2$, and so $\displaystyle x=1$.
Can also be done by taking logs to the base 5:
$\displaystyle
(x+1)\log_5(5)=2\log_5(5)
$
but $\displaystyle \log_5(5)=1$, so we are back to where we were, and $\displaystyle x=1$
RonL
5a. Simplify:
$\displaystyle
\log_2(5)+\log_2(x^2-1)-\log_2(x-1)
$.
Here the first things we need are some of the basic laws of logarithms:
$\displaystyle \log_a(u)+\log_s(v)=\log_a(u\times v)$,
and:
$\displaystyle \log_a(u)-\log_s(v)=\log_a(u/v)$.
Now:
$\displaystyle
\log_2(5)+\log_2(x^2-1)$$\displaystyle -\log_2(x-1)=(\log_2(5)+\log_2(x^2-1))-\log_2(x-1)
$
$\displaystyle
=\log_2(5(x^2-1))-\log_2(x-1)
$$\displaystyle
=\log \left(\frac{5(x^2-1}{x-1}\right)
$
Finally we need to recall how to factorise the difference of two squares to get:
$\displaystyle
=\log \left(\frac{5(x-1)(x+1)}{x-1}\right)=\log(5(x+1))
$
RonL