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3 years ago

Find the equation of the line normal to the curve of y=cos1/3x, where x=pi

Mathematics

1 answer:

miv72 [106K]3 years ago

5 0

Answer:

The equation of the line normal to the curve of y=cos1/3x, where x=pi is y= 2√3 x - 10.38

Step-by-step explanation:

The equation of the normal to the curve is

Y- y1 = - 1/m (x- x1)

Where m is the gradient.

See attached picture for the complete solution.

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C(1)=−20 c(n)=c(n−1)+10 Find the second term in the sequence.

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Answer:

c(2) = -10

Step-by-step explanation:

The first equation says that the first term of the sequence is -20.

The second equation is saying that to find any term of the sequence, add 10 to the previous term.

c(2) = c(2-1) + 10

c(2) = c(1) + 10

c(2) = -20 + 10 = -10

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Read 2 more answers

Tell whether the lines through the given points are parallel, perpendicular, or neither?

Paha777 [63]

$Slope\,\,1=\frac{-1}{slope\,\,2}$

So, Both lines are perpendicular

Step-by-step explanation:

We need to identify if lines:

Line 1: (8,12)(7,-5)

Line 2: (-9,3)(8,2)

are parallel, perpendicular, or neither?

Lines are parallel if they have same slopes i.e slope 1= slope2
Lines are perpendicular if their slopes are negative inverses of each other i.e slope 1= -1/slope 2

The formula used to find slope is:

$Slope=\frac{y_2-y_1}{x_2-x_1}$

Finding Slope of Line 1:

The points are: Line 1: (8,12)(7,-5)

$x_1=8, y_1=12, x_2=7,y_2=-5$

Putting values in the formula:

$Slope\,\,1=\frac{y_2-y_1}{x_2-x_1}$

$Slope\,\,1=\frac{-5-(12)}{7-8}$

$Slope\,\,1=\frac{-5-12}{-1}$

$Slope\,\,1=\frac{-17}{-1}$

$Slope\,\,1=17$

So, Slope of line 1 is 17

Finding Slope of Line 2:

The points are: Line 2: (-9,3)(8,2)

$x_1=-9, y_1=3, x_2=8,y_2=2$

Putting values in the formula:

$Slope\,\,2=\frac{y_2-y_1}{x_2-x_1}$

$Slope\,\,2=\frac{2-(3)}{8-(-9)}$

$Slope\,\,2=\frac{2-3}{8+9}$

$Slope\,\,2=\frac{-1}{17}$

So, Slope of line 2 is $\frac{-1}{17}$

Since $Slope\,\,1=\frac{-1}{slope\,\,2}$

So, Both lines are perpendicular

Keywords: Slope of Lines

Learn more about Slope of lines at:

#learnwithBrainly

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