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Marysya12
3 months ago
14

ABCD is a quadrilateral inscribed in a circle, as shown below: Circle O is shown with a quadrilateral ABCD inscribed inside it.

Angle A is labeled x plus 16. Angle B is labeled x degrees. Angle C is labeled 6x minus 4. Angle D is labeled 2x plus 16. What equation can be used to solve for angle C? (x + 16) + (x) = 180 (x + 16) + (6x − 4) = 180 (x + 16) − (2x + 16) = 180 (x + 16) + (x) = 90
Mathematics
1 answer:
PIT_PIT [12.4K]3 months ago
5 0

Answer:

B. (x + 16) + (6x − 4) = 180

Step-by-step explanation:

The Theorem of Cyclic Quadrilaterals states that in any quadrilateral located within a circle, the angles opposite to each other will sum to 180 degrees. Therefore, the conclusion can be drawn:

∠A + ∠C = 180° leading to (x + 16) + (6x - 4) = 180°

∠B + ∠D = 180° which equates to x + (2x - 16) = 180°

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Answer:

in steps

Step-by-step explanation:

In a fair roulette game, the likelihood of the ball landing on RED remains constant, regardless of prior spins.

a) 18/38

b) 18/38

c) Yes, I have confidence in my responses. Since it's fair, the number of RED slots does not change.

8 0
1 month ago
Describe the sample space for each of these experiments (a) A coin is tossed and a single die is rolled. (b) A paper cup is toss
PIT_PIT [12445]

Answer:

a) S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) S={top,bottom,side}

c) S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

Step-by-step explanation:

The sample space consists of all potential outcomes for each experiment:

a) The sample encompasses the outcomes resulting from tossing a coin (heads or tails) and throwing a die (a number between 1 and 6):

S={(head,1),(head,2),(head,3),(head,4),(head,5),(head,6),(tails,1),(tails,2),(tails,3),(tails,4),(tails,5),(tails,6)}

b) A paper cup can settle in one of three different positions: on its top, bottom, or side:

S={top,bottom,side}

c) The entire set of cards within a standard deck constitutes the sample space:

S={(Ace of diamonds),(Two of diamonds),(Three of diamonds),(Four of diamonds),(Five of diamonds),(Six of diamonds),(Seven of diamonds), (Eight of diamonds),(Nine of diamonds),(Ten of diamonds), (Jack of diamonds), (Queen of diamonds),(King of diamonds),(Ace of hearts),(Two of hearts),(Three of hearts),(Four of hearts),(Five of hearts),(Six of hearts),(Seven of hearts), (Eight of hearts), (Nine of hearts),(Ten of hearts), (Jack of hearts), (Queen of hearts),(King of hearts),(Ace of clubs),(Two of clubs),(Three of clubs),(Four of clubs),(Five of clubs),(Six of clubs),(Seven of clubs), (Eight of clubs), (Nine of clubs),(Ten of clubs), (Jack of clubs),(Queen of clubs),(King of clubs),(Ace of spades),(Two of spades),(Three of spades),(Four of spades),(Five of spades),(Six of spades),(Seven of spades),(Eight of spades), (Nine of spades),(Ten of spades),(Jack of spades),(Queen of spades),(King of spades)}

3 0
2 months ago
Which fraction is equivalent to StartFraction 2 Over 6 EndFraction? StartFraction 3 Over 7 EndFraction, because StartFraction 2
babunello [11817]
Which equivalent fraction corresponds to \frac{2}{6}? - \frac{3}{7} since \frac{2}{6}= \frac{2 + 1}{6 + 1} - \frac{3}{9} as \frac{2}{6}= \frac{1}{3} and \frac{1}{3} = \frac{3}{9}

- \frac{3}{12} because \frac{2}{6}= \frac{1}{3} and \frac{1}{3} = \frac{3}{9}

- \frac{3}{8} since \frac{2}{6}= \frac{1}{2} = \frac{2+1}{6+2} = \frac{3}{8}

The answer is

- \frac{3}{9} because \frac{2}{6}= \frac{1}{3} and \frac{1}{3} = \frac{3}{9}

A fraction is deemed equivalent if it maintains the same value when expressed in simplest terms. The equivalent to \frac{2}{6} is found in the chosen option;

First, halve both the numerator and denominator by 2

\frac{2/2}{6/2}

Then reduce further

2/2 = 1 and 6/2 = 3; Thus;

\frac{2/2}{6/2} = \frac{1}{3}

Next, multiply both the numerator and denominator by 3

\frac{1*3}{3*3} = \frac{3}{9}

Therefore, \frac{2}{6} equals \frac{3}{9}.

7 0
2 months ago
Read 2 more answers
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