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Sonbull [250]
2 years ago
9

Which of the following are adjacent angles?

Mathematics
1 answer:
sleet_krkn [62]2 years ago
6 0

Answer:

Solution :

********************************************

Two angles are said to be a pair of

adjacent angles if they have a common

arm and lie on the either sides of the

common arm .

***********************************************

Now ,

i ) a , b are a pair of adjecent angles .

ii ) c , d are pair of adjecent angles.

iii ) e, f are not adjecent angles .

Step-by-step explanation:

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Point E is where the statue should be placed.

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At what point should the statue be:

The intersection of the doors to the arena, the spirit store, and the parking lot forms a triangle, with Point E in its center. The radius of the circumscribed circle will therefore equal the distance from the statue to each vertex, placing it at the same distance from all three landmarks.

Point E is where the statue should be placed.

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4 0
1 year ago
ZAP is a right triangle, with right angle A and cosP = 4/5. What is sin Z?
irga5000 [103]
If you want to find cosP , since angles Z and P are 2 complementary angles then cosZ is the same as cosP so you say the cosP=sinZ=4/5 (property of 2 complement angles in a right triangle)
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3 years ago
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Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco
kari74 [83]

Answer:

C. \frac{1}{18}

Step-by-step explanation:

Given: Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is 8, what is the probability that one of the two cards drawn is numbered 5.

Solution:

Sample space for sum of cards when two cards are drawn at random is \{(1,1),(1,2),(1,3)......(6,6)\}

total number of possible cases =36

Sample space when sum of cards is 8 is \{(3,5),(5,3),(6,2),(2,6),(4,4)\}

Total number of possible cases =5

Sample space when one of the cards is 5 is \{(5,3),(3,5)\}

Total number of possible cases =2

Let A be the event that sum of cards is 8

p(\text{A}) =\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}

p(\text{A})=\frac{5}{36}

Let B be the event when one of the two cards is 5

probability than one of two cards is 5 when sum of cards is 8

p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}

p(\frac{\text{B}}{\text{A}})=\frac{2}{5}

Now,

probability that sum of cards 8 is and one of cards is 5

p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})

p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}

p(\text{A and B})=\frac{1}{18}

if sum of cards is 8 then probability that one of the cards is 8 is \frac{1}{18}, option C is correct.

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Drag and drop the number to match the division problem to its quotient.
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C I'm using the calculator and the other ones don't make sense

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293 seconds ____________
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