The complete question is
<span>Sarah rides her horse for 2 1/6 hrs with a constant speed of 6 km/hr and then for another 2 hours and 5 minutes with a constant speed of 12 km/hr what is her average speed for the total trip?</span>
we know that
speed=distance/time
1) Sarah rides her horse for 2 1/6 hrs with a constant speed of 6 km/hr
time=2 1/6 hrs----> 13/6 hrs
speed=6 km/h
distance=speed*time----> 6*(13/6)---> 13 km
2)for another 2 hours and 5 minutes with a constant speed of 12 km/hr
time=2hrs+5 minutes----> 2+(5/60)--> 125/60 hrs
speed=12 km/hrs
distance=speed*time----> 12*(125/60)---> 25 km
3) find the average speed for the total trip
total distance=13 km+25 km----> 38 km
total time=(13/6)+(125/60)----> (10*13+125)/60----> 4.25 hours
speed=38/4.25----> 8.94 km/hrs
the answer is
8.94 km/hrs
⇒I will first isolate y together with its coefficient k by placing
to the right hand side...

⇒Now to leave y independent we have to divide ky by the coefficient of y which in this case is k.
⇒Meaning k will divide all the terms in the equation.

⇒Attached is the answer.
Zeroes
set numerator to zero
x^2=0
x=0
zeroes are at (0,0)
HA
for p(x)/q(x)
when degree of p(x)<q(x), HA=0
when degree of p(x)=q(x), HA= leading
coef of p(x) divided by leading coef of q(x)
when degree of p(x)>q(x) you
probably have a slant assymtote
degrees are same (no slant assymtote)
leading coefs
1/1=1
HA is y=1
does it cross?
1=(x^2)/(x^2-4)
x^2-4=x^2
minus x^2 both sides
-4=0
false, it does not cross the HA
VA
simplify the fraction (it can't)
set the denomenator to zero
(x-2)(x+2)=0
VA's at x=2 and x=-2
so to graph
graph the point (0,0)
draw the lines
y=1
x=-2
x=2
reemmber, plus, minus, plus
so from left, it goes from above the HA
right up to the VA of x=-2
then goes upside down U shape in
between VA's going through (0,0)
then from top of VA x=2 down to y=1
then gets closer to y=1 but never touches