What are similarities between the stories Trifles and “A Jury of Her Peers”?

# What are similarities between the stories Trifles and “A Jury of Her Peers”?

• MATH

C=43∘,a=49,b=79C=43∘,a=49,b=79 Use the Law of Cosines to solve the triangle. Round your…

Using the law of cosines, in the given case, yields: c2=a2+b2−2ab⋅cosCc2=a2+b2-2ab⋅cosC c2=(49)2+(79)2−2⋅289⋅cos43oc2=(49)2+(79)2-2⋅289⋅cos43o c2=6581−5681⋅7310c2=6581-5681⋅7310 c2=(181)⋅650−408810c2=(181)⋅650-408810

• MATH

C=15∘15′,a=7.45,b=2.15C=15∘15′,a=7.45,b=2.15 Use the Law of Cosines to solve the triangle. Round your…

Given: C=15∘15,a=7.45,b=2.15C=15∘15,a=7.45,b=2.15 c2=a2+b2−2ab⋅cos(C)c2=a2+b2-2ab⋅cos(C) c2=7.452+2.152−2(7.45)(2.15)cos(15.25)c2=7.452+2.152-2(7.45)(2.15)cos(15.25) c2=29.2280c2=29.2280 c=5.41c=5.41 cos(A)=b2+c2−a22bccos(A)=b2+c2-a22bc…

• MATH

B=125∘40′,a=37,c=37B=125∘40′,a=37,c=37 Use the Law of Cosines to solve the triangle. Round your…

Given B=125∘40′,a=37,c=37B=125∘40′,a=37,c=37 b2=a2+c2−2ac⋅cos(B)b2=a2+c2-2ac⋅cos(B) b2=372+372−2(37)(37)cos(125∘40′)b2=372+372-2(37)(37)cos(125∘40′) b2=4334.44b2=4334.44 b=65.68b=65.68 Since side a and side c are congruent, triangles ABC is an isosceles…

• MATH

B=75∘20′,a=6.2,c=9.5B=75∘20′,a=6.2,c=9.5 Use the Law of Cosines to solve the triangle. Round your…

Given: B=75∘20′,a=6.2,c=9.5B=75∘20′,a=6.2,c=9.5 b2=a2+c2−2ac⋅cos(B)b2=a2+c2-2ac⋅cos(B) b2=6.22+9.52−2(6.2)(9.5)cos(75∘20′)b2=6.22+9.52-2(6.2)(9.5)cos(75∘20′) b2=98.8636b2=98.8636 b=9.94b=9.94 cos(C)=a2+b2−c22abcos(C)=a2+b2-c22ab…

• MATH

B=10∘35′,a=40,c=30B=10∘35′,a=40,c=30 Use the Law of Cosines to solve the triangle. Round your…

Given: B=10∘35′,a=40,c=30B=10∘35′,a=40,c=30 b2=a2+c2−2ac⋅cos(B)b2=a2+c2-2ac⋅cos(B) b2=402+302−2(40)(30)cos(10.583)b2=402+302-2(40)(30)cos(10.583) b2=140.824b2=140.824 b=11.87b=11.87 cos(A)=b2+c2−a22bccos(A)=b2+c2-a22bc cos(A)=11.872+302−4022⋅11.87⋅30cos(A)=11.872+302-4022⋅11.87⋅30…

• MATH

A=48∘,b=3,c=14A=48∘,b=3,c=14 Use the Law of Cosines to solve the triangle. Round your answers…

Given: A=48∘,b=3,c=14A=48∘,b=3,c=14  a2=b2+c2−2bc⋅cos(A)a2=b2+c2-2bc⋅cos(A) a2=32+142−2⋅3⋅14⋅cos(48)a2=32+142-2⋅3⋅14⋅cos(48) a2=148.79a2=148.79 a=12.20a=12.20  cos(C)=a2+b2−c22⋅a⋅bcos(C)=a2+b2-c22⋅a⋅b cos(C)=12.202+32−1422⋅12.20⋅3cos(C)=12.202+32-1422⋅12.20⋅3 …

• MATH

A=120∘,b=6,c=7A=120∘,b=6,c=7 Use the Law of Cosines to solve the triangle. Round your answers…

The given in the triangle are A=120oA=120o , b=6b=6 and c=7c=7 . To solve for the values of a, B and C, let’s apply Cosine Law. For side a: a2=b2+c2−2⋅b⋅c⋅cosAa2=b2+c2-2⋅b⋅c⋅cosA a2=62+72−2⋅6⋅7cos(120o)a2=62+72-2⋅6⋅7cos(120o)…

• MATH

a=1.42,b=0.75,c=1.25a=1.42,b=0.75,c=1.25 Use the Law of Cosines to solve the triangle. Round your…

Given: a=1.42,b=0.75,c=1.25a=1.42,b=0.75,c=1.25 cos(A)=b2+c2−a22bccos(A)=b2+c2-a22bc cos(A)=0.752+1.252−1.4222⋅0.75⋅1.25cos(A)=0.752+1.252-1.4222⋅0.75⋅1.25 cos(A)=.05792cos(A)=.05792 A=arccos(.05792)A=arccos(.05792) A=86.68∘A=86.68∘ cos(C)=a2+b2−c22abcos(C)=a2+b2-c22ab…

• MATH

a=75.4,b=52,c=52a=75.4,b=52,c=52 Use the Law of Cosines to solve the triangle. Round your answers…

Given: a=75.4,b=52,c=52a=75.4,b=52,c=52  cos(A)=b2+c2−a22bccos(A)=b2+c2-a22bc  cos(A)=522+522−7522⋅52⋅52cos(A)=522+522-7522⋅52⋅52 cos(A)=−.04012cos(A)=-.04012 A=arccos(−.04012)A=arccos(-.04012) A=92.30∘A=92.30∘ cos(B)=a2+c2−b22accos(B)=a2+c2-b22ac …

• MATH

a=55,b=25,c=72a=55,b=25,c=72 Use the Law of Cosines to solve the triangle. Round your answers to…

Given: a=55,b=25,c=72a=55,b=25,c=72  cos(C)=a2+b2−c22abcos(C)=a2+b2-c22ab  cos(C)=552+252−7222⋅55⋅25cos(C)=552+252-7222⋅55⋅25 cos(C)=−.5578cos(C)=-.5578 C=arccos(−.5578)C=arccos(-.5578) C=123.91∘C=123.91∘ cos(A)=b2+c2−a22bccos(A)=b2+c2-a22bc…

• MATH

a=11,b=15,c=21a=11,b=15,c=21 Use the Law of Cosines to solve the triangle. Round your answers to…

Given: a=11,b=15,c=21a=11,b=15,c=21  cos(C)=a2+b2−c22abcos(C)=a2+b2-c22ab cos(C)=112+152−2122⋅11⋅15cos(C)=112+152-2122⋅11⋅15 cos(C)=−.2878cos(C)=-.2878 C=arccos(−.2878)C=arccos(-.2878) C=106.73∘C=106.73∘ cos(B)=a2+c2−b22accos(B)=a2+c2-b22ac…

• THE BROTHERS KARAMAZOV

How is the theme of children portrayed in the Brothers of Karamazov?

What an interesting idea for a simplistic theme for this very deep Russian novel. I agree that “children” could definitely be a very simple theme in The Brothers Karamazov by Fyodor Dostoevsky,…

• MATH

C=84∘30′,a=16,b=20C=84∘30′,a=16,b=20 Find the area of the triangle having the indicated angle and…

Given: C=84∘30′,a=16,b=20C=84∘30′,a=16,b=20 Use the Area formula A=12(a)(b)sin(C)A=12(a)(b)sin(C) C=84∘30’=84.5∘C=84∘30′=84.5∘ A=12(16)(20)sin(84.5)A=12(16)(20)sin(84.5) A=159.26A=159.26 The area is 159.26 squared units.

• MATH

B=72∘30′,a=105,c=64B=72∘30′,a=105,c=64 Find the area of the triangle having the indicated angle and…

Given: B=72∘30′,a=105,c=64B=72∘30′,a=105,c=64 Use the Area formula A=12(a)(c)sin(B)A=12(a)(c)sin(B) B=72∘30’=72.5∘B=72∘30′=72.5∘ A=12(105)(64)sin(72.5)A=12(105)(64)sin(72.5) A=3204.49A=3204.49 The area is 3204.49 squared units.

• MATH

A=5∘15′,b=4.5,c=22A=5∘15′,b=4.5,c=22 Find the area of the triangle having the indicated angle and…

Given: A=5∘15′,b=4.5,c=22A=5∘15′,b=4.5,c=22 Use the Area formula A=12(b)(c)sin(A)A=12(b)(c)sin(A) A=5∘15’=5.25∘A=5∘15′=5.25∘ A=12(4.5)(22)sin(5.25)A=12(4.5)(22)sin(5.25) A=4.53A=4.53 The area is 4.53 squared units.

• MATH

A=43∘45′,b=57,c=85A=43∘45′,b=57,c=85 Find the area of the triangle having the indicated angle and…

Given: A=43∘45′,b=57,c=85A=43∘45′,b=57,c=85 Use the Area Formula A=12(b)(c)sin(A)A=12(b)(c)sin(A) A=43∘45’=43.75∘A=43∘45′=43.75∘ A=12(57)(85)sin(43.75)A=12(57)(85)sin(43.75) A=1675.19A=1675.19 The area is 1675.19 squared units.

• MATH

C=170∘,a=14,b=24C=170∘,a=14,b=24 Find the area of the triangle having the indicated angle and sides.

Given: C=170∘,a=14,b=24C=170∘,a=14,b=24 Use the Area Formula A=12(a)(b)sin(C)A=12(a)(b)sin(C) A=12(14)(24)sin(170)A=12(14)(24)sin(170) A=29.17A=29.17 The area is 29.17 squared units.

• MATH

A=150∘,b=8,c=10A=150∘,b=8,c=10 Find the area of the triangle having the indicated angle and sides.

Given A=150∘,b=8,c=10A=150∘,b=8,c=10 Use the Area Formula A=12bcsin(A)A=12bcsin(A) A=12(8)(10)sin(150)A=12(8)(10)sin(150) Area is 20 square units.

• MATH

B=130∘,a=62,c=20B=130∘,a=62,c=20 Find the area of the triangle having the indicated angle and sides.

Given B∘=130,a=62,c=20B∘=130,a=62,c=20 Use the Area formula A=12(a)(c)sin(B∘)A=12(a)(c)sin(B∘) A=12(62)(20)sin(130)A=12(62)(20)sin(130) A=474.95A=474.95 The area is 474.95 squared units.

• MATH

C=120∘,a=4,b=6C=120∘,a=4,b=6 Find the area of the triangle having the indicated angle and sides.

Given C=120∘,a=4,b=6C=120∘,a=4,b=6 Use the Area formula A=12(a)(b)sin(C)A=12(a)(b)sin(C) A=12(4)(6)sin(120)A=12(4)(6)sin(120) A=10.39A=10.39 The area is 10.39 square units.

• MATH

A=25∘4′,a=9.5,b=22A=25∘4′,a=9.5,b=22 Use the law of sines to solve (if possible) the triangle. If…

Given: A=25∘4′,a=9.5,b=22A=25∘4′,a=9.5,b=22 Law of Sines asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 9.5sin(25.6666)=22sin(B)=csin(C)9.5sin(25.6666)=22sin(B)=csin(C)9.5sin(25.6666)=22sin(B)9.5sin(25.6666)=22sin(B) sin(B)=22sin(25.6666)9.5sin(B)=22sin(25.6666)9.5 sin(B)=1.003sin(B)=1.003…

• MATH

A=45∘,a=b=1A=45∘,a=b=1 Use the law of sines to solve (if possible) the triangle. If two…

Given: A=45∘,a=1,b=1A=45∘,a=1,b=1 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 1sin(45)=1sin(B)=csin(C)1sin(45)=1sin(B)=csin(C) 1sin(45)=1sin(B)1sin(45)=1sin(B) sin(B)=sin(45)sin(B)=sin(45) B=45∘B=45∘ C=180−45−45C=180-45-45 C=90∘C=90∘…

• MATH

A=120∘,a=25,b=24A=120∘,a=25,b=24 Use the law of sines to solve (if possible) the triangle. If…

Given: A=120∘,a=25,b=24A=120∘,a=25,b=24 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 25sin(120)=24sin(B)=csin(C)25sin(120)=24sin(B)=csin(C) sin(B)=24sin(120)25=.8313sin(B)=24sin(120)25=.8313 B=arcsin(.8313)B=arcsin(.8313)  B=56.24B=56.24 C=180−120−56.24C=180-120-56.24…

• MATH

A=120∘,a=b=25A=120∘,a=b=25 Use the law of sines to solve (if possible) the triangle. If two…

The given in the triangle are A=120^o, a=25 and b=25. To solve using Sine Law, let’s use the formula: sinAa=sinBbsinAa=sinBb Plug-in the values of the sides a and b. Also, plug-in the value of angle…

• MATH

A=58∘,a=4.5,b=12.8A=58∘,a=4.5,b=12.8 Use the law of sines to solve (if possible) the triangle. If…

Given: A=58∘,a=4.5,b=12.8A=58∘,a=4.5,b=12.8 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 4.5sin(58)=12.8sin(B)=csin(C)4.5sin(58)=12.8sin(B)=csin(C) 4.5sin(58)=12.8sin(B)4.5sin(58)=12.8sin(B) sin(B)=12.8sin(58)4.5sin(B)=12.8sin(58)4.5 sin(B)=2.4122sin(B)=2.4122Angle B…

• MATH

A=58∘,a=11.4,b=12.8A=58∘,a=11.4,b=12.8 Use the law of sines to solve (if possible) the triangle. If…

Given: A=58∘,a=11.4,b=12.8A=58∘,a=11.4,b=12.8 Law of Sines asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) Triangle #1 11.4sin(58)=12.8sin(B)=csin(C)11.4sin(58)=12.8sin(B)=csin(C) 11.4sin(58)=12.8sin(B)11.4sin(58)=12.8sin(B) sin(B)=12.8sin(58)11.4sin(B)=12.8sin(58)11.4 sin(B)=.9522sin(B)=.9522…

• MATH

A=76∘,a=34,b=21A=76∘,a=34,b=21 Use the law of sines to solve (if possible) the triangle. If two…

Given: A=76∘,a=34,b=21A=76∘,a=34,b=21 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 34sin(76)=21sin(B)=csin(C)34sin(76)=21sin(B)=csin(C) 34sin(76)=21sin(B)34sin(76)=21sin(B) sin(B)=21sin(76)34sin(B)=21sin(76)34 sin(B)=.5993sin(B)=.5993 B=arcsin(.5993)B=arcsin(.5993)…

• MATH

A=76∘,a=18,b=20A=76∘,a=18,b=20 Use the law of sines to solve (if possible) the triangle. If two…

Given: A=76∘,a=18,b=20A=76∘,a=18,b=20 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 18sin(76)=20sin(B)=csin(C)18sin(76)=20sin(B)=csin(C) 18sin(76)=20sin(B)18sin(76)=20sin(B) sin(B)=20sin(76)18sin(B)=20sin(76)18 sin(B)=1.0781sin(B)=1.0781 B=arcsin(1.0781)B=arcsin(1.0781)…

• MATH

A=110∘,a=125,b=200A=110∘,a=125,b=200 Use the law of sines to solve (if possible) the triangle. If…

Given: A=100∘,a=125,b=200A=100∘,a=125,b=200 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 125sin(100)=200sin(B)=csin(C)125sin(100)=200sin(B)=csin(C)  125sin(100)=200sin(B)125sin(100)=200sin(B) sin(B)=200sin(100)125sin(B)=200sin(100)125 sin(B)=1.57sin(B)=1.57…

• MATH

A=110∘,a=125,b=100A=110∘,a=125,b=100 Use the law of sines to solve (if possible) the triangle. If…

Given: A=110∘,a=125,b=100A=110∘,a=125,b=100 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 125sin(110)=100sin(B)=csin(C)125sin(110)=100sin(B)=csin(C) 125sin(110)=100sin(B)125sin(110)=100sin(B) sin(B)=100sin(110)125sin(B)=100sin(110)125 sin(B)=.7518sin(B)=.7518…

• MATH

C=95.20∘,a=35,c=50C=95.20∘,a=35,c=50 Use the law of sines to solve the triangle. (Find missing…

Given: C=95.20∘,a=35,c=50C=95.20∘,a=35,c=50 Law of Sines asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 35sin(A)=bsin(B)=50sin(95.20)35sin(A)=bsin(B)=50sin(95.20) 35sin(A)=50sin(95.20)35sin(A)=50sin(95.20) sin(A)=35sin(95.20)50sin(A)=35sin(95.20)50 sin(A)=.6971sin(A)=.6971 A=arcsin(.6971)A=arcsin(.6971)…

• MATH

A=110∘15′,a=48,b=16A=110∘15′,a=48,b=16 Use the law of sines to solve the triangle. (Find missing…

Given: A=110∘15′,a=48,b=16A=110∘15′,a=48,b=16 Law of Sines asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 48sin(110.25)=16sin(B)=csin(C)48sin(110.25)=16sin(B)=csin(C) 48sin(110.25)=16sin(B)48sin(110.25)=16sin(B) sin(B)=16sin(110.25)48sin(B)=16sin(110.25)48 sin(B)=.3127sin(B)=.3127…

• MATH

A=100∘,a=125,c=10A=100∘,a=125,c=10 Use the law of sines to solve the triangle. (Find missing…

Given: A=100∘,a=125,c=10A=100∘,a=125,c=10 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 125sin(100)=bsin(B)=10sin(C)125sin(100)=bsin(B)=10sin(C) sin(C)=10sin(100)125sin(C)=10sin(100)125 sin(C)=.0788sin(C)=.0788 C=arcsin(.0788)C=arcsin(.0788) C=4.52∘C=4.52∘…

• MATH

A=145∘,a=14,b=4A=145∘,a=14,b=4 Use the law of sines to solve the triangle. (Find missing…

Given: A=145∘,a=14,b=4A=145∘,a=14,b=4 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) 14sin(145)=4sin(B)=csin(C)14sin(145)=4sin(B)=csin(C) 14sin(145)=4sin(B)14sin(145)=4sin(B) sin(B)=4sin(145)14sin(B)=4sin(145)14 sin(B)=.1639sin(B)=.1639 B=arcsin(.1639)B=arcsin(.1639)…

• MATH

B=2∘45′,b=6.2,c=5.8B=2∘45′,b=6.2,c=5.8 Use the law of sines to solve the triangle. (Find missing…

The sine law indicates that in a triangle; sinAa=sinBb=sinCcsinAa=sinBb=sinCca,b,c are the lengths of the triangle and A,B,C are the opposite angles as shown in image. B=2045’=2.750B=2045′=2.750 Using…

• MATH

B=15∘30′,a=4.5,b=6.8B=15∘30′,a=4.5,b=6.8 Use the law of sines to solve the triangle. (Find missing…

The sine law indicates that in a triangle; sinAa=sinBb=sinCcsinAa=sinBb=sinCca,b,c are the lengths of the triangle and A,B,C are the opposite angles as shown in image. Angle A =15030’=15.50=15030′=15.50…

• MATH

A=60∘,a=9,b=5A=60∘,a=9,b=5 Use the law of sines to solve the triangle. (Find missing…

The sine law indicates that in a triangle; sinAa=sinBb=sinCcsinAa=sinBb=sinCca,b,c are the lengths of the triangle and A,B,C are the opposite angles as shown in image. sinAa=sinBbsinAa=sinBb

• MATH

A=36∘,a=8,b=5A=36∘,a=8,b=5 Use the law of sines to solve the triangle. (Find missing…

The sine law indicates that in a triangle; sinAa=sinBb=sinCcsinAa=sinBb=sinCca,b,c are the lengths of the triangle and A,B,C are the opposite angles as shown in image. sinAa=sinBbsinAa=sinBb

• MATH

B=28∘,C=104∘,a=358B=28∘,C=104∘,a=358 Use the law of sines to solve the triangle. (Find missing…

Given: B=28∘,C=104∘,a=3.625B=28∘,C=104∘,a=3.625 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) A=180−28−104=48∘A=180-28-104=48∘ 3.625sin(48)=bsin(28)=csin(104)3.625sin(48)=bsin(28)=csin(104) 3.625sin(48)=bsin(28)3.625sin(48)=bsin(28) b=3.625sin(28)sin(48)b=3.625sin(28)sin(48)…

• MATH

A=55∘,B=42∘,c=34A=55∘,B=42∘,c=34 Use the law of sines to solve the triangle. (Find missing…

Given: A=55∘,B=42∘,c=0.75A=55∘,B=42∘,c=0.75 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) C=180−55−42=83∘C=180-55-42=83∘ asin(55)=bsin(42)=0.75sin(83)asin(55)=bsin(42)=0.75sin(83) asin(55)=0.75sin(83)asin(55)=0.75sin(83) a=0.75sin(55)sin(83)a=0.75sin(55)sin(83)…

• MATH

A=120∘,B=45∘,c=16A=120∘,B=45∘,c=16 Use the law of sines to solve the triangle. (Find missing…

Given: A=120∘,B=45∘,c=16A=120∘,B=45∘,c=16 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) C=180−120−45=15∘C=180-120-45=15∘ asin(120)=bsin(45)=16sin(15)asin(120)=bsin(45)=16sin(15) asin(120)=16sin(15)asin(120)=16sin(15) a=16sin(120)sin(15)=53.54a=16sin(120)sin(15)=53.54…

• MATH

A=35∘,B=65∘,c=10A=35∘,B=65∘,c=10 Use the law of sines to solve the triangle. (Find missing…

Given: A=35∘,B=65∘,c=10A=35∘,B=65∘,c=10 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C)  C=180−35−65=80C=180-35-65=80 asin(35)=bsin(65)=10sin(80)asin(35)=bsin(65)=10sin(80) asin(35)=10sin(80)asin(35)=10sin(80) a=10sin(35)sin(80)a=10sin(35)sin(80) a=5.82a=5.82…

• MATH

A=5∘40′,B=8∘15′,b=4.8A=5∘40′,B=8∘15′,b=4.8 Use the law of sines to solve the triangle. (Find…

Using law of sines sinAa=sinBb=sinCcsinAa=sinBb=sinCc A,B and C are angles and a,b and c are opposite lengths of a triangle as in the image. According to the question we have following data.

• MATH

A=83∘20′,C=65.6∘,c=18.1A=83∘20′,C=65.6∘,c=18.1 Use the law of sines to solve the triangle. (Find…

Using law of sines sinAa=sinBb=sinCcsinAa=sinBb=sinCc A,B and C are angles and a,b and c are opposite lengths of a triangle as in the image. According to the question we have following data.

• MATH

A=24.3∘,C=54.6∘,c=2.68A=24.3∘,C=54.6∘,c=2.68 Use the law of sines to solve the triangle. (Find…

Given: A=24.3∘,C=54.6∘,c=2.68A=24.3∘,C=54.6∘,c=2.68 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C) B=180−24.3−54.6=101.1∘B=180-24.3-54.6=101.1∘ asin(24.3)=bsin(101.1)=2.68sin(54.6)asin(24.3)=bsin(101.1)=2.68sin(54.6) asin(24.3)=2.68sin(54.6)asin(24.3)=2.68sin(54.6)…

• MATH

A=102.4∘,C=16.7∘,a=21.6A=102.4∘,C=16.7∘,a=21.6 Use the law of sines to solve the triangle. (Find…

Given A=102.4∘,C=16.7∘,a=21.6A=102.4∘,C=16.7∘,a=21.6 The Law of Sines: asin(A)=bsin(B)=csin(C)asin(A)=bsin(B)=csin(C)  B=180−102.4−16.7=60.9∘B=180-102.4-16.7=60.9∘  21.6sin(102.4)=bsin(60.9)=csin(16.7)21.6sin(102.4)=bsin(60.9)=csin(16.7) 21.6sin(102.4)=bsin(60.9)21.6sin(102.4)=bsin(60.9)…

• HISTORY

What two guiding principles held by the Framers most affected the structure and functions of the…

The first guiding principle shared by most of the framers was the need for a powerful central government. James Madison in particular had come to believe that the states, if left to their own…

• THE LIGHT IN THE FOREST

Why does Half Arrow watch quietly while the guard binds True Son’s arms in Chapter 4 of The Light…

In the story, John Butler was only four years old when he was abducted by Delaware Indians. Renamed True Son by his adopted Indian father, Cuyloga, John comes to love and to respect the Indian way…

• A JURY OF HER PEERS

What are similarities between the stories Trifles and “A Jury of Her Peers”?

Susan Glaspell is successfully able to transform her play Trifles into what would become her most famous story, “A Jury of Her Peers”. In reality, it is easier to extrapolate the differences…