Given that ΔJKL ≅ ΔUVW and ΔUVW ≅ ΔABC, complete the following statements. Triangle JKL is congruent to triangle ABCBCACBA. Side
2 answers:
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Answer:
The answer is "$ 11,961.43"
Step-by-step explanation:
Given values:
P = $ 9500
r= 2.1 %
time (t)= 11 year
total quarterly year =4
Formula:
quarterly time =
Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA⊥PA
OB⊥PB
In △OPA and △OPB
∠OPA=∠OPB (Using (1))
OA=OB (Radii of the same circle)
OP=OP (Common side)
Therefor △OPA≅△OPB (RHS congruency criterion)
PA=PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
The length of tangents drawn from any external point are equal.
So statement is correct
To prove: PA = PB
Construction: Join OA, OB, and OP.
It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.
OA⊥PA
OB⊥PB
In △OPA and △OPB
∠OPA=∠OPB (Using (1))
OA=OB (Radii of the same circle)
OP=OP (Common side)
Therefor △OPA≅△OPB (RHS congruency criterion)
PA=PB
(Corresponding parts of congruent triangles are equal)
Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.
The length of tangents drawn from any external point are equal.
So statement is correct
Answer:
Step-by-step explanation:
given is the Differential equation in I order linear as
Take Laplace on both sides
Now if we take inverse we get y(t) the solution
Thus the algebraic equation would be