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3 years ago

Angle G is a circumscribed angle of circle E. Major arc FD measures 280°.

Mathematics

2 answers:

mixer [17]3 years ago

8 0

Answer:

The answer is 40°

Step-by-step explanation:

Solution

Now,

The Measure of major arc FD = 280°

Thus,

One complete angle measure 360°

Then

m∠FED = 360 - 180

m∠FED = 80°

Thus,

FE = DE ( the same circle Radius)

∠EFD = ∠EDF ( opposite angles to equal sides are equal)

Now

Applying angle sum property of a triangle in ΔEFD

∠EFD + ∠EDF + ∠FED = 180°

∠EFD + ∠EFD + 80 = 180

2∠EFD = 100

∠EFD = 50°

Hence, GF is tangent to the circle and the tangent always make right angles with the radius of the circle.

∠EFG = 90°

∠GFD + ∠EFD = 90°

∠GFD + 50 = 90

∠GFD = 40°

Therefore, The measure of the angle GFD is 40°

MAXImum [283]3 years ago

5 0

Answer:

Step-by-step explanation:

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Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

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$=2^03^{-8}$

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Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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