We now have an algebraic expression with only one variable, which can be solved. Once we have that, we can plug it back into one
of the original equations (or the expression derived in Part A) to solve for the other variable. When this is done with the system of two equations from Parts A and B, what is the solution?
1 answer:
Answer:
A=1
B=-2
Explanation:
Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.
From part B as attached, it shows that the right option is C which is
2A+3B=-4
Substituting B with 3A-5 then we form the second equation as shown
2A+3(3A-5)=-4
By simplifying the above equation, we obtain
2A+9A-15=-4
Re-arranging, then
11A=-4+15
Finally
11A=11
A=1
To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain
B=3(1)-5=-2
Therefore, required values are 1 and -2
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Answer:
Explanation:
The general equation for position of Simple harmonic motion is given as:
........(1)
where,
x = Position of the wave
A = Amplitude of the wave
ω = Angular velocity
t = time
In this case, the amplitude is just half the range,
thus,
(Given range = 3cm)
A = 1.5 cm
Now, The angular velocity is given as:
Where, T = time period of the wave =0.27s (given)
or
so, at time t = 55 s, the equation (1) becomes as:
on solving the above equation we get,
here the negative sign depicts the position in the opposite direction of +x