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coldgirl [10]
3 years ago
9

Indicate the equation of the line meeting the given conditions. Please put the equation in standard form. Containing A (1, 3) an

d B (0, 2)
Mathematics
1 answer:
wolverine [178]3 years ago
3 0
A line through two points A, B can be found by the point-slope form of the formula:

(y-ya)=m(x-xa)............(1)
and 
m=(yb-ya)/(xb-xa).......(2)

where A(xa,ya), B(xb,yb) and m is the slope between points A & B.

Substituting
A(xa,ya)=A(1,3)
B(xb,yb)=B(0,2)

From (2)
m=(yb-ya)/(xb-xa)
=(2-3)/(0-1)
=-1/(-1)
=1

Substitute in (1)  : (y-ya)=m(x-xa)
y-3=1(x-1)
Distribute and simplify
y=x-1+3=x+2
or
y=x+2  .................(3a) equation required in slope-intercept form

x-y+2=0...............(3b) equation in standard form
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5 shares if only whole shares are allowed

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Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt
Paladinen [302]

Answer:

y(x)=4a\sqrt{3}* sin(x)-3a

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}

But:

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So:

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Multiplying both sides by dx and dividing both sides by 3a+y:

\frac{dy}{3a+y} =cot(x)dx

Integrating both sides:

\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx

Evaluating the integrals:

log(3a+y)=log(sin(x))+C_1

Where C1 is an arbitrary constant.

Solving for y:

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So:

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Therefore:

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