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3 years ago

Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mathematics

1 answer:

Mumz [18]3 years ago

4 0

Answer:

$Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

Step-by-step explanation:

The equation of the curve is

$Y = sin^{-1}(7x)$

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

$Y'=\frac{7}{\sqrt{1-49x^2} }$

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is $x = \sqrt{\frac{1}{7} }$

Then slope would accordingly be

$Y'=\frac{7}{\sqrt{1-49/49} }$

= ∞

For, $x = \sqrt{\frac{1}{7} }$ , $Y = sin^{-1}(7/7)= \pi/2$

Equation of tangent line, in the point slope form, would be $Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

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Bezzdna [24]

Answer:

The answer is c = 2 + 3d .

Step-by-step explanation:

It is given that a cab ride cost $2 which is a fixed amount and charges additional $3 per mile. So you have to make an equation of c in terms of d :

Let c be the cost,

Let d be the distance,

c = 2 + 3d

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3 years ago

The measures of the three angles of a triangle are given by 1. 3x +1 2. 2x -3, and 9x. What is the measure of the smallest angle

nordsb [41]

Since the total measure of all the angles of a triangle is always 180, solve for x with the equation: 3x+1+2x-3+9x=180. x=13. Now plug in x and look for the smallest one, which will be B. 23 degrees.

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3 years ago

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Tanzania [10]

Answers:

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3 years ago

Find all Values of b that will make the polynomial a perfect square trinomial.

weeeeeb [17]

Answer:

10,404/334,084

Step-by-step explanation:

Given the polynomial

289r^2 - 102r + c

We are to find the value of c that will make it a perfect square

Divide through by 289

289r²/289 - 102r/289 + c/289

Half of the coefficient of r is 1/2(102/289)

Half of the coefficient of r = 102/578

Square the result

r² = (102/578)²

r² = 10,404/334,084

Hence the required constant is 10,404/334,084

8 0

3 years ago

SOMEONE HELP MEEEEEE 75 POINTS TO THE PERSON THAT HELPS

Tresset [83]

Answer:

Part 1) $9x-7y=-25$

Part 2) $2x-y=2$

Part 3) $x+8y=22$

Part 4) $x+8y=35$

Part 5) $3x-4y=2$

Part 6) $10x+6y=39$

Part 7) $x-5y=-6$

Part 8)

case A) The equation of the diagonal AC is $x+y=0$

case B) The equation of the diagonal BD is $x-y=0$

Step-by-step explanation:

Part 1)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M=(\frac{2-6}{2},\frac{-3+5}{2})$

$M=(-2,1)$

step 2

The equation of the line into point slope form is equal to

$y-1=\frac{9}{7}(x+2)\\ \\y=\frac{9}{7}x+\frac{18}{7}+1\\ \\y=\frac{9}{7}x+\frac{25}{7}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=\frac{9}{7}x+\frac{25}{7}$

Multiply by 7 both sides

$7y=9x+25$

$9x-7y=-25$

Part 2)

step 1

Find the midpoint

The formula to calculate the midpoint between two points is equal to

$M=(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M=(\frac{1+5}{2},\frac{0-2}{2})$

$M=(3,-1)$

step 2

Find the slope

The slope between two points is equal to

$m=\frac{-2-0}{5-1}=-\frac{1}{2}$

step 3

we know that

If two lines are perpendicular, then the product of their slopes is equal to -1

Find the slope of the line perpendicular to the segment joining the given points

$m1=-\frac{1}{2}$

$m1*m2=-1$

therefore

$m2=2$

step 4

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=2$ and point $(1,0)$

$y-0=2(x-1)\\ \\y=2x-2$

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=2x-2$

$2x-y=2$

Part 3)

In this problem AB and BC are the legs of the right triangle (plot the figure)

step 1

Find the midpoint AB

$M1=(\frac{-5+1}{2},\frac{5+1}{2})$

$M1=(-2,3)$

step 2

Find the midpoint BC

$M2=(\frac{1+3}{2},\frac{1+4}{2})$

$M2=(2,2.5)$

step 3

Find the slope M1M2

The slope between two points is equal to

$m=\frac{2.5-3}{2+2}=-\frac{1}{8}$

step 4

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{8}$ and point $(-2,3)$

$y-3=-\frac{1}{8}(x+2)\\ \\y=-\frac{1}{8}x-\frac{1}{4}+3\\ \\y=-\frac{1}{8}x+\frac{11}{4}$

step 5

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=-\frac{1}{8}x+\frac{11}{4}$

Multiply by 8 both sides

$8y=-x+22$

$x+8y=22$

Part 4)

In this problem the hypotenuse is AC (plot the figure)

step 1

Find the slope AC

The slope between two points is equal to

$m=\frac{4-5}{3+5}=-\frac{1}{8}$

step 2

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=-\frac{1}{8}$ and point $(3,4)$

$y-4=-\frac{1}{8}(x-3)$

$y=-\frac{1}{8}x+\frac{3}{8}+4$

$y=-\frac{1}{8}x+\frac{35}{8}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=-\frac{1}{8}x+\frac{35}{8}$

Multiply by 8 both sides

$8y=-x+35$

$x+8y=35$

Part 5)

The longer diagonal is the segment BD (plot the figure)

step 1

Find the slope BD

The slope between two points is equal to

$m=\frac{4+2}{6+2}=\frac{3}{4}$

step 2

The equation of the line into point slope form is equal to

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$ and point $(-2,-2)$

$y+2=\frac{3}{4}(x+2)$

$y=\frac{3}{4}x+\frac{6}{4}-2$

$y=\frac{3}{4}x-\frac{2}{4}$

step 3

Convert to standard form

Remember that the equation of the line into standard form is equal to

$Ax+By=C$

where

A is a positive integer, and B, and C are integers

$y=\frac{3}{4}x-\frac{2}{4}$

Multiply by 4 both sides

$4y=3x-2$

$3x-4y=2$

Note The complete answers in the attached file

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3 0

3 years ago