The possible values of b are 7 and 8
<h3>How to determine the possible values of x?</h3>
The expression is given as:
4x² + bx + 3
Next, we test the options to determine the values of b
<u>Option 1: b = 13</u>
So, we have:
4x² + 13x + 3 ---- this cannot be factorized
<u>Option 2: b = 7</u>
So, we have:
4x² + 7x + 3
Expand
4x² + 4x + 3x + 3
Factorize
4x(x +1) + 3(x + 1)
Factor out x + 1
(4x + 3)(x + 1) ------ this can be factorized
<u>Option 3: b = 8</u>
So, we have:
4x² + 8x + 3
Expand
4x² + 6x + 2x + 3
Factorize
2x(2x +3) + 1(2x + 3)
Factor out 2x + 3
(2x + 3)(2x + 1) ------ this can be factorized
<u>Option 4: b = 1</u>
So, we have:
4x² + x + 3 ---- this cannot be factorized
Hence, the possible values of b are 7 and 8
Read more about factorized expressions at:
brainly.com/question/723406
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1] ⬇
p - (6 - 2(q))➡0
p + 2q - 6➡0
q = 3 because 6÷2 = 0
p = 6 because 6 ÷ 1 = 6
Slope ➡ 0.5
-----------------------------------------------------------------
2] ⬇
p - (1 + q)➡0
p - q- 1➡0
Q = -1 because 1 ÷ -1 ='-1
P = 1 because 1 ÷ 1 = 1
Slope ➡ 1
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