Answer:
Option C. is correct
Step-by-step explanation:
In the triangle ABC, vertices are A(12, 8), B(4, 8) and C(4, 14)
AB =
BC =
AC =
In triangle XYZ vertices are X(6,6), Y(4,12) and Z(10, 14)
XY =
YZ =
XZ =
In triangle MNO, vertices are M(4, 16), N(4, 8) and O(-2, 8)
MN =
MO =
NO =
In triangle JKL, vertices are J(14, -2), K(12,2) and L(20, 4)
JK =
KL =
JL =
Now we compare the sides for the congruence
Since AB = MN = 8
BC = NO = 6
AC = OM = 10
So ΔABC ≅ Δ MNO
Option C. is correct
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Answer:
quadrant 3
Step-by-step explanation:
Answer:
a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)
Step-by-step explanation:
Let's solve by separating variables:

a) x’=t–sin(t), x(0)=1

Apply integral both sides:

where k is a constant due to integration. With x(0)=1, substitute:

Finally:

b) x’+2x=4; x(0)=5

Completing the integral:

Solving the operator:

Using algebra, it becomes explicit:

With x(0)=5, substitute:

Finally:

c) x’’+4x=0; x(0)=0; x’(0)=1
Let
be the solution for the equation, then:

Substituting these equations in <em>c)</em>

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>
![x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)](https://tex.z-dn.net/?f=x%3De%5E%7B%5Calpha%20t%7D%5BAsin%5Cbeta%20t%2BBcos%5Cbeta%20t%5D%5C%5C%5C%5Cx%3De%5E%7B0%7D%5BAsin%28%282%29t%29%2BBcos%28%282%29t%29%5D%5C%5C%5C%5Cx%3DAsin%28%282%29t%29%2BBcos%28%282%29t%29)
Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

Finally:

Answer:
i belive its 6 please tell me if im wrong
Step-by-step explanation:
Okay I’ll solve this but it’ll take a while to answer but I’ll try my best to finish as fast as I can!