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Doss [256]
3 years ago
6

What is the equation of a line, in point- slope form, that passes through (5, -3) and has a slope of 2/3?

Mathematics
1 answer:
Alika [10]3 years ago
8 0
<h3><u><em>Option B</em></u></h3><h3><u><em>The equation of a line, in point- slope form, that passes through (5, -3) and has a slope of 2/3</em></u></h3><h3><u><em /></u>y + 3 = \frac{2}{3}(x -5)<u><em /></u></h3>

<em><u>Solution:</u></em>

<em><u>The equation of line in point slope form is given as:</u></em>

y - y_1 = m(x-x_1)

Where, "m" is the slope of line

Given that,

slope = 2/3

m = \frac{2}{3}

The line passes through (5, -3)

Substitute m = 2/3 and (x_1, y_1) = (5, -3) in point slope form

y - (-3) = \frac{2}{3}(x -5)\\\\y + 3 = \frac{2}{3}(x -5)

Thus equation of line in point slope form is found

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3 years ago
Find the general solution of the following ODE: y' + 1/t y = 3 cos(2t), t &gt; 0.
Margarita [4]

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

3 0
3 years ago
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