Answer:
Step-by-step explanation:
Given the simultaneous equation 2p - 3q = 4 and 3p + 2q = 9, to get the value of p and q we will use elimination method.
2p - 3q = 4 ...................... 1 * 3
3p + 2q = 9 ..................... 2 * 2
Multiplying equation 1 by 3 and 3 by 2:
6p - 9q = 12
6p + 4q = 18
Subtracting both equation
-9q-4q = 12-18
-13q = -6
q = -6/-13
q = 6/13
Substituting q = 6/13 into equation 2
2p - 3(6/13) = 4
2p - 18/13 = 4
2p = 4+18/13
2p = (52+18)/13
2p = 70/13
p = 70/26
p = 35/13
<em>Hence p = 35/13 and q = 6/13</em>
<em></em>
<em>b) </em>If if 223ₓ = 87 find x
Using the number base system and converting 223ₓ to base 2 will give us;
223ₓ = 2*x² + 2*x¹ + 3*x⁰
223ₓ = 2x²+2x+3
Substituting back into the equation, 2x²+2x+3 = 87
2x²+2x+3-87 = 0
2x²+2x-84 = 0
x²+x-42 = 0
On factorizing:
(x²+6x)-(7x-42) = 0
x(x+6)-7(x+6) = 0
(x+6)(x-7) = 0
x+6 = 0 and x-7 = 0
x = -6 and 7
<em>Hence the value of x is 7 (neglecting the negative value)</em>
Answer:
y = 3sin2t/2 - 3cos2t/4t + C/t
Step-by-step explanation:
The differential equation y' + 1/t y = 3 cos(2t) is a first order differential equation in the form y'+p(t)y = q(t) with integrating factor I = e^∫p(t)dt
Comparing the standard form with the given differential equation.
p(t) = 1/t and q(t) = 3cos(2t)
I = e^∫1/tdt
I = e^ln(t)
I = t
The general solution for first a first order DE is expressed as;
y×I = ∫q(t)Idt + C where I is the integrating factor and C is the constant of integration.
yt = ∫t(3cos2t)dt
yt = 3∫t(cos2t)dt ...... 1
Integrating ∫t(cos2t)dt using integration by part.
Let u = t, dv = cos2tdt
du/dt = 1; du = dt
v = ∫(cos2t)dt
v = sin2t/2
∫t(cos2t)dt = t(sin2t/2) + ∫(sin2t)/2dt
= tsin2t/2 - cos2t/4 ..... 2
Substituting equation 2 into 1
yt = 3(tsin2t/2 - cos2t/4) + C
Divide through by t
y = 3sin2t/2 - 3cos2t/4t + C/t
Hence the general solution to the ODE is y = 3sin2t/2 - 3cos2t/4t + C/t
Answer:
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Answer:
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Step-by-step explanation:
