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12345 [234]
3 years ago
5

OC bisects AOB, OD bisects AOC, OE bisects AOD, OF bisects AOE, OG bisects BOF. if BOF =120°, find DOE

Mathematics
1 answer:
LiRa [457]3 years ago
8 0

Answer:

∠DOE = 16°

Step-by-step explanation:

The given parameters are;

∠BOF = 120°

∠AOB = 2×∠AOC                        {}      Given

∠AOC = 2×∠AOD                     {}         Given

∠AOD = 2×∠AOE      {}                         Given

∠AOE = 2×∠AOF {}                               Given

Therefore;

∠AOB = 16×∠AOF                 {}              Angle addition postulate

∠BOF = ∠AOB - ∠AOF = 16×∠AOF  - ∠AOF = 15×∠AOF {} Transitive property

15×∠AOF = 120°  

∠AOF = 120°/15 = 8°

Given that OE bisects ∠AOD, we have;

∠AOE ≅ ∠DOE                      {}                    Angles bisected by a line

From;

∠AOE = 2×∠AOF, we have;  {}                               Given

Therefore;

∠AOE = ∠DOE = 2×∠AOF = 2×8° = 16°

∠DOE = 16°.

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The graph below represents the solution set of which inequality?
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option: B (x^2+2x-8) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

x^2-2x-8

On using the method of splitting the middle term we have:

x^2-4x+2x-8

⇒  x(x-4)+2(x-4)=0

⇒ (x+2)(x-4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

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x+2>0 and x-4

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

x+2 and x-4>0

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

x^2+2x-8

On using the method of splitting the middle term we have:

x^2+4x-2x-8

⇒ x(x+4)-2(x+4)

⇒ (x-2)(x+4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x-2>0 and x+4

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

x-2 and x+4>0

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

x^2-2x-8>0

On using the method of splitting the middle term we have:

x^2-4x+2x-8>0

⇒ x(x-4)+2(x-4)>0

⇒ (x-4)(x+2)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x+2>0 and x-4>0

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

x+2 and x-4

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

x^2+2x-8>0

On using the method of splitting the middle term we have:

x^2+4x-2x-8>0

⇒ x(x+4)-2(x+4)>0

⇒ (x-2)(X+4)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x-2 and x+4

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

x-2>0 and x+4>0

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




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HELPPPPP IM GIVING 50 POINTS FOR THIS !
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16 cups = 4 quarts

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