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Nostrana [21]
3 years ago
15

Circle G is inscribed with triangle E F D. Point C is on the circle between points E and F. Angle E is 79 degrees. The measure o

f arc E D is 104 degrees. What is the measure of arc ECF in circle G? 52° 98° 158° 177°

Mathematics
2 answers:
melomori [17]3 years ago
7 0

Answer:

Option B.

Step-by-step explanation:

Given information: Circle G is inscribed with triangle EFD, arcED=104°.

We need to find the measure of arc ECF.

Draw a diagram by the given information.

Arc(ED)=104^{\circ}              (Given)

Arc(FD)=2(\angle E)=2(79)=158^{\circ}            (Central angle theorem)

Arc(ECF)=360-Arc(ED)-Arc(FD)

Arc(ECF)=360-104-158

Arc(ECF)=98^{\circ}

The measure of arc EDF is 98°.

Therefore, the correct option is B.

steposvetlana [31]3 years ago
3 0

Answer:

98

Step-by-step explanation:

79x2

158+104=360

262=360

=98

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a. The equation of the tangent at (1,3) is y = -x/3 + 10/3

b. The equation of the normal at (1,3) is y = 3x

c. Find the graph in the attachment

<h3>a. How to find the equation of the tangent at (1, 3)?</h3>

Since x² + y² = 10, we differentiate the equation with respect to x to find dy/dx which the the equation of the tangent.

So, x² + y² = 10

d(x² + y²)/dx = d10/dx

dx²/dx + dy²/dx = 0

2x + 2ydy/dx = 0

2ydy/dx = -2x

dy/dx = -2x/2y

dy/dx = -x/y

At (1,3), dy/dx = -1/3

Using the equation of a straight line in slope-point form, we have

m = (y - y₁)/(x - x₁) where

  • m = gradient of the tangent = dy/dx at (1,3) = -1/3 and
  • (x₁, y₁) = (1,3)

So, m = (y - y₁)/(x - x₁)

-1/3 = (y - 3)/(x - 1)

-(x - 1) = 3(y - 3)

-x + 1 = 3y - 9

3y = -x + 1 + 9

3y = -x + 10

3y + x = 10

y = -x/3 + 10/3

So, the equation of the tangent at (1,3) is y = -x/3 + 10/3

<h3>b. The equation of the normal at the point (1, 3)</h3>

Since the tangent and normal line are perpendicular at the point, for two perpendicular line,

mm' = -1 where

  • m = gradient of tangent = -1/3 and
  • m' = gradient of normal

So, m' = -1/m

= -1/(-1/3)

= 3

Using the equation of a straight line in slope-point form, we have

m' = (y - y₁)/(x - x₁) where

  • m' = gradient of normal at (1, 3)  and (
  • x₁, y₁) = (1,3)

So, m = (y - y₁)/(x - x₁)

3 = (y - 3)/(x - 1)

3(x - 1) = (y - 3)

3x - 3 = y - 3

y = 3x - 3 + 3

y = 3x + 0

y = 3x

So, the equation of the normal at (1,3) is y = 3x

c. Find the graph in the attachment

Learn more about equation of tangent and normal here:

brainly.com/question/7252502

#SPJ1

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