<span>Consider a angle â BAC and the point D on its defector
Assume that DB is perpendicular to AB and DC is perpendicular to AC.
Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC
Proof
Consider triangles ΔADB and ΔADC
Both are right angle, â ABD= â ACD=90 degree
They have congruent acute angle â BAD and â CAD( since AD is angle bisector)
They share hypotenuse AD
therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
Pedro scored 1479 points while Ricky scored 1199 points
Answer with explanation:
The given differential equation is
y" -y'+y=2 sin 3x------(1)
Let, y'=z
y"=z'
Substituting the value of , y, y' and y" in equation (1)
z'-z+zx=2 sin 3 x
z'+z(x-1)=2 sin 3 x-----------(1)
This is a type of linear differential equation.
Integrating factor
Multiplying both sides of equation (1) by integrating factor and integrating we get
Answer:
53/21 OR 2 11/21
Step-by-step explanation:
1 4/6 + 6/7
Change 1 4/6 into an improper fraction
1 4/6 6x1= 6+4= 10/6
Simplify 10/6 --> 5/3
5/3+ 6/7
find a common denominator (21)
35/21+18/21= 53/21
mixed number form is 2 11/21
