<ol>
<li><p>Let the function f be defined by f(x)=x squared +18. If m is a positive number such that f(2m)=2 f(m)
what is the value of m?</p></li>
<li><p>figure shown is a parrallelogram ABCD with line DB drawn in across the figure dividing into two triangles.
If the five line segments in the figure above are all congruent, what is the ratio of the length of AC (not shown) to the length of BD?</p></li>
</ol>
<p>a. square root of 2 to 1
b. square root of 3 to 1
c. square root of2 to 2
d. square root of 3 to 2
e. square root of 3 to square root of 2</p>
<ol>
<li><p>If n and p are integers greater than 1 and is p is a factor of both n+3 and n+10, then what is the value of p?
a.3
b.7
c.10
d13
e 30</p></li>
<li><p>If xy=7 and x-y=5 then x squared y -x y squared =</p></li>
</ol>
<p>1] The first is very easy:
f(2m)=(2m)^2+18=4m^2+18
2f(m)=2(x^2+18)=2x^2+36
we look for m such as
2f(m)=f(2m) or 4m^2+18=2x^2+36 (I put a lot of step to be clear, OK?)
2m^2-18=0
2(m^2-9)=0
2(m-3)(m+3)=0 So, there are two solutions: -3 and 3</p>
<p>2) According to the informations provided, your parallelogram is actually a rhombus or two equilateral triangles sharing a same side.
Now, If you can’t visualize, draw a schema and everything will be translucent.
Let a be the side of our two triangles.
the ratio is (lenght of AC)/(lenght of DB)
DB=a
and AC=2<em>altitude. altitude is equal to a</em>sqrt(3)/2 (You can prove it easily thanks to pytagorean theorem)
So AC= a*sqrt(3)</p>
<p>In the ratio, a/a=1 (Obvious?) so you have your answer: sqrt of 3 to 1</p>
<p>3) p is a factor of n+3 means that p<em>x=n+3
Likewise p</em>y=n+10 (x and y different, OK)
As they are related let’s substract them</p>
<p>p<em>y-p</em>x=n+10-(n+3) —>n+10-n-3 If you had forgotten
p<em>y-p</em>x=7
7 is a prime number and has only two factors 1 and 7
But p>1 so p=7</p>
<p>I hope I am clear (I’m an international), BTW could you make 4) more clear, I mean it isn’t structured (brackets???)
<p>I realize that I made a lot of mistake. It was night (time difference).
“more clear”–>clearer and so on… SHAME ON ME!
BTW
I forgot to tell you that in an equilateral triangle, altitudes, bisectors, angle bisectors and medians are the same lines.
You could thereby replace “altitude” by any of them above</p>