25-4x=16-4x
25-4x+4x=16-4x+4x
25=16
No solution
Contradiction
Answer: hello your question is poorly written below is the complete question
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
answer:
a ) R is equivalence
b) y = 2x + C
Step-by-step explanation:
<u>a) Prove that R is an equivalence relation </u>
Every line is seen to be parallel to itself ( i.e. reflexive ) also
L1 is parallel to L2 and L2 is as well parallel to L1 ( i.e. symmetric ) also
If we presume L1 is parallel to L2 and L2 is also parallel to L3 hence we can also conclude that L1 is parallel to L3 as well ( i.e. transitive )
with these conditions we can conclude that ; R is equivalence
<u>b) show the set of all lines related to y = 2x + 4 </u>
The set of all line that is related to y = 2x + 4
y = 2x + C
because parallel lines have the same slopes.
Answer: Darren
A) Job Offer #1 would be better financially until job offer 2 (in 12 years) would be a better and have more income, so IMO job offer #1 would set you better off financially
B) it would take 12 years for job offer #2 to become a better offer (you can also express this in your own words if needed)
Answer:
y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1
Step-by-step explanation:
y" + y' + y = 1
This is a second order nonhomogenous differential equation with constant coefficients.
First, find the roots of the complementary solution.
y" + y' + y = 0
r² + r + 1 = 0
r = [ -1 ± √(1² − 4(1)(1)) ] / 2(1)
r = [ -1 ± √(1 − 4) ] / 2
r = -1/2 ± i√3/2
These roots are complex, so the complementary solution is:
y = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t)
Next, assume the particular solution has the form of the right hand side of the differential equation. In this case, a constant.
y = c
Plug this into the differential equation and use undetermined coefficients to solve:
y" + y' + y = 1
0 + 0 + c = 1
c = 1
So the total solution is:
y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1
To solve for c₁ and c₂, you need to be given initial conditions.
