A.
( 2.5 , 11 )
( 5 , 11 )
( 4.5 , 8 )
( 2 , 8 )
( 7 , 11 )
( 7 , 8 )
( 12 , 11 )
( 4.5 , 6 )
( 4.5 , 2 )
( 2 , 2 )
( 2 , 6 )
( 8 , 2 )
( 8 , 6 )
( 11 , 1 )
( 11 , 7 )
Answer:
maybe I think the answer is 50
Answer: read the comments for extra added more accurate information this was in the comments and by HOZI3RSMUS3 so it’s not mine
We're suppose to measure the trigonometric values of the triangles and the values of indicates angles .For that,We need to know that,Sin θ = side opposite to angle θ / HypotenuseCos θ = side adjacent to angle θ / HypotenuseTan θ = side opposite to angle θ/side adjacent to angle θNow,In the first triangle,We're provided by,side opposite to angle θ =QR =24side adjacent to angle θ = PR = 10Hypotenuse = PQ = 26So,Sin P = 24/26Cos P = 10/26Tan P = 24/10m∠P = sin p° = 24/26m∠P= sin^-1(24/26)m∠P= 67.38 °In the second triangle,We're provided by,side opposite to angle θ =MO =9side adjacent to angle θ = NO = 40Hypotenuse = MN= 41So,Sin N= 9/41Cos N= 40/41Tan N= 9/40m∠N = sin N° = 9/41m∠N= sin^-1(9/41)m∠N= 12.68°In the third triangle,We're provided by,side opposite to angle θ =AC =15side adjacent to angle θ = AB = 8Hypotenuse = BC = 17So,Sin B = 15/17Cos B = 8/17Tan B = 15/8m∠B = sin p° = 15/17m∠B= sin^-1(15/17)m∠B= 61.93°In the fourth triangle,We're provided by,side opposite to angle θ =BC =8side adjacent to angle θ = AC = 6Hypotenuse = AB = 10So,Sin A = 8/10Cos A = 6/10Tan A = 8/6m∠A = sin p° = 8/10m∠A= sin^-1(8/10)m∠P= 53. °*53.13°
Step-by-step explanation:
Answer:
Step-by-step explanation:
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Collecting like terms and expanding the brackets
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Answer:
domain is like the membrane
Step-by-step explanation:
domain
