It depends, if the two sides given are adjacent to each other then yes. but you are given two opposite sides then no, because you don't know the length of one of the sides, thus not being able to calculate the perimeter
Answer:
+
*LN(|
|) +C
Step-by-step explanation:
we will have to do a trig sub for this
use x=a*tanθ for sqrt(x^2 +a^2) where a=2
x=2tanθ, dx= 2 sec^2 (θ) dθ
this turns
into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ
the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1
then it simplifies into integral(4*sec^3 (θ)) dθ
you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C
then you will need to rework your functions of θ back into functions of x
tanθ will resolve back into
(see substitutions) while secθ will resolve into
sec(θ)=
is from its ratio identity of hyp/adj where the hyp. is
and adj is 2 (see tan(θ) ratio)
after resolving back into functions of x, substitute ratios for trig functions:
=
+
*LN(|
|) +C


-3 is our y-intercept is plotted on both graphs yet one is negative and the other is positive and

Is our slope and is positive and the graph that is positive is the first graph.
Answer:
a vertical asymptote at x=-4
Step-by-step explanation:

The binomial denominator factor is not canceled by a numerator factor, so represents a vertical asymptote at x=-4. That is the only discontinuity