We know from Euclidean Geometry and the properties of a centroid that GC=2GM. Now GM=sin60*GA=
. Hence CM=GC+GM=3*
. Now, since GM is normal to AB, we have by the pythagoeran theorem that:
Hence, we calculate from this that AC^2= 28, hence AC=2*
=BC. Thus, the perimeter of the triangle is 2+4*
.
$150 because the supply number is equivalent to the demand number
Answer: 0
Step-by-step explanation: hope it helps
Answer:heyyy there...the answer is b that is -12x^3+12x^2-20
Step-by-step explanation:
<h3>f(x)=
-12x^3+19x^2-5</h3><h3>g(x)=
7x^2+15</h3><h3>f(x)-g(x)=(
-12x^3+19x^2-5)-( 7x^2+15)...{while opening the bracket the sign of the second polynomial changes accordingly}</h3><h3>
it becomes -12x^3+19x^2-5-7x^2-15</h3><h3>
=-12x^3+12x^2-20</h3><h3>
HOPE IT HELPED UUUU</h3>
Take the homogeneous part and find the roots to the characteristic equation:
This means the characteristic solution is
.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With
and
, you're looking for a particular solution of the form
. The functions
satisfy
where
is the Wronskian determinant of the two characteristic solutions.
So you have
So you end up with a solution
but since
is already accounted for in the characteristic solution, the particular solution is then
so that the general solution is
