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Tanya [424]
2 years ago
15

Write the equation of a line that is parallel to {y=0.6x+3}y=0.6x+3y, equals, 0, point, 6, x, plus, 3 and that passes through th

e point {(-3,-5)}(−3,−5)left parenthesis, minus, 3, comma, minus, 5, right parenthesis.
Mathematics
1 answer:
Dominik [7]2 years ago
3 0

y = 0.6x - 3.2  
The general form of the desired equation is 
y = mx + b
 where
 m = slope of the line
 b = y intercept of the line 
 If two lines are parallel, their slopes will be the same, Since the slope of the
given line "y = 0.6x +3" is 0.6, that will also be the slope of the desired line.
So our equation becomes:
 y = 0.6x + b 
 Now we can substitute the x and y value of the desired point we want the new line to pass through and find b. So
 y = 0.6x + b
 -5 = 0.6(-3) + b
 -5 = -1.8 + b
 -3.2 = b 
 So the desired equation is now 
y = 0.6x - 3.2
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Step-by-step explanation:

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<em><u>Correct the error. Write the correct decomposition:</u></em>

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5 0
3 years ago
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Select the curve generated by the parametric equations. Indicate with an arrow the direction in which the curve is traced as t i
bixtya [17]

Answer:

length of the curve = 8

Step-by-step explanation:

Given parametric equations are x = t + sin(t) and y = cos(t) and given interval is

−π ≤ t ≤ π

Given data the arrow the direction in which the curve is traces means

the length of the curve of the given parametric equations.

The formula of length of the curve is

\int\limits^a_b {\sqrt{\frac{(dx}{dt}) ^{2}+(\frac{dy}{dt}) ^2 } } \, dx

Given limits values are −π ≤ t ≤ π

x = t + sin(t) ...….. (1)

y = cos(t).......(2)

differentiating equation (1)  with respective to 'x'

\frac{dx}{dt} = 1+cost

differentiating equation (2)  with respective to 'y'

\frac{dy}{dt} = -sint

The length of curve is

\int\limits^\pi_\pi  {\sqrt{(1+cost)^{2}+(-sint)^2 } } \, dt

\int\limits^\pi_\pi  \,   {\sqrt{(1+cost)^{2}+2cost+(sint)^2 } } \, dt

on simplification , we get

here using sin^2(t) +cos^2(t) =1 and after simplification , we get

\int\limits^\pi_\pi  \,   {\sqrt{(2+2cost } } \, dt

\sqrt{2} \int\limits^\pi_\pi  \,   {\sqrt{(1+1cost } } \, dt

again using formula, 1+cost = 2cos^2(t/2)

\sqrt{2} \int\limits^\pi _\pi  {\sqrt{2cos^2\frac{t}{2} } } \, dt

Taking common \sqrt{2} we get ,

\sqrt{2}\sqrt{2}  \int\limits^\pi _\pi ( {\sqrt{cos^2\frac{t}{2} } } \, dt

2(\int\limits^\pi _\pi  {cos\frac{t}{2} } \, dt

2(\frac{sin(\frac{t}{2} }{\frac{t}{2} } )^{\pi } _{-\pi }

length of curve = 4(sin(\frac{\pi }{2} )- sin(\frac{-\pi }{2} ))

length of the curve is = 4(1+1) = 8

<u>conclusion</u>:-

The arrow of the direction or the length of curve = 8

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Answer:

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Step-by-step explanation:

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