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2

Answer:

# color(purple)( A ~~ 89.2941 #

Explanation:

Problem:
Use Heron's to determine the area of a triangle with sides 15, 18, and 12 units long.

Heron's formula:
# color(blue)( A = sqrt( s(s - a)(s - b)(s - c) )#, and
# color(blue)( s = (a + b + c)/(2)#.

Calculate s and the stuff inside the radical.
# s = ((15) + (18) + (12))/(2) #
# s = (45)/(2) #.
# s - a = 45/2 - 30/2 = 15/2 #

# s - b = 45/2 - 36/2 = 9/2 #

# s - c = 45/2 - 24/2 = 21/2 #

# s(s - a)(s - b)(s - c) = 45/2 * 15/2 * 9/2 * 21/2 #

# s(s - a)(s - b)(s - c) = 127575/16 #

Now, the area.
# A = sqrt( 127575/16 ) #
# color(purple)( A ~~ 89.2941 #

1

Answer:

#tan(B/4)=(sqrt13-3)/2#

Explanation:

Here,

#a=5 , b=12 and c=13#

#=>BC=5, CA=12 ,AB=13#

#=>bar(AB) # , is the hypotenuse of #triangleABC#

So, #mangleC=90^circ=>angle A and angle B# are acute angles.

enter image source here

Now, #color(red)(cosB=("side adjacent to B")/("hypotenus")=5/13#

Using half angle formula :

#cos(B/2)=(1+cosB)/2=(1+5/13)/2=9/13=(3/sqrt13)^2#

#=>color(red)(cos(B/2)=+3/sqrt13...to[because mangleB < 90^circ]#

Also ,we know the half angle formula:

#color(blue)(tan^2(theta/2)=(1-costheta)/(1+costheta)#

Let us take, #theta=(B/2)#

#:.color(blue)(tan^2((B/2)/2)=(1-cos(B/2))/(1+cos(B/2))#

#=>tan^2(B/4)=(1-3/sqrt13)/(1+3/sqrt13)=(sqrt13-3)/(sqrt13+3)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((1)#

#=>tan^2(B/4)=(sqrt13-3)/(sqrt13+3)xxcolor(green)((sqrt13-3)/(sqrt13-3)#

#=>tan^2(B/4)=(sqrt13-3)^2/((sqrt13)^2-(3)^2)#

#=>tan^2(B/4)=(sqrt13-3)^2/4#

#=>(tan(B/4))^2=((sqrt13-3)/2)^2...tomangleB <90^circ#

#=>tan(B/4)=(sqrt13-3)/2#

1

Answer:

More than needed.

Explanation:

In the sentence "This is a single unpaired sock" there is redundancy in the form of twice saying the same (also known as tautology).

A person can be "made redundant", meaning he is no longer needed for a certain job at a certain office or factory (there are too many in that job at that place).

Calling a person redundant (overall) would not be a very nice thing to say: It would mean the person is good for nothing and might as well die.

1

Answer:

Amerigo Vespucci was a Florentine explorer and navigator who first determined that the east coast of North Amercia was part of a seperate landmass from Asia.

Explanation:

"America" largely results from the Latin version of Americao Vespucci's first name. Born in Florence in 1454, he was originally a merchant and entrepreneur but started to serve both the Portuguese and Spanish governments as a navigator.

In 1497-1504 Vespucci determined that the coast of South America extended much further than Cristopher Columbus had thought, and argued (in a widely circulated series of letters) that this meant South America might be a major landmass that was not connected to Asia.

Like many of his contemporaries (and other explorers), Vespucci was a little bit of a huckster, and some of his accomplishments on his four voyages may be questionable. Moreover, subsequent claims by others added to his lustre -- partly due to a scholarly debate over whether it was Vespucci or Columbus who truely discovered America.

Regardless, it remains that Columbus believed his discoveries were outlying parts of Asia, but Vespucci argued -- successfully that there was a major landmass, a continent in its own right, to the west of Europe and the east of Asia. Europe's leading cartographer Martin Waldseemüller named this land mass "America" in Vespucci's honor. Vespucci died in 1512.

2

Answer:

because is it 2 different kinds of media combined

Explanation:

whenever text, audio, still images, video, animation etc. are combined, the result is multimedia.
think of multimedia as multiple medias.

1

Answer:

#16S^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=(15+18+12)(-15+18+12)(15-18+12)(15+18-12)=(45)(15)(9)(21)=3(15)(15)(9)(3)(7)# or

#S ={ 9 (15)}/4 sqrt{7}={135 sqrt{7} }/4#

Explanation:

Seems a bit easier than the other answer.

2

#tanx/(1-cotx) + cotx/(1-tanx) = -1#

#=>-tanx/(1-1/tanx) - (1/tanx)/(1-tanx) = 1#

#=>-tan^2x/(tanx-1) - 1/(tanx(1-tanx) )= 1#

#=>tan^2x/(1-tanx) - 1/(tanx(1-tanx) )= 1#

#=>(tan^3x - 1)/(tanx(1-tanx) )= 1#

#=>((tanx - 1)(tan^2x+tanx+1))/(tanx(1-tanx) )= 1#

#tanx=1# will make LHS undefined so neglected.
Hence

#=> - (tan^2x+tanx+1)/tanx= 1#

#=> tan^2x+2tanx+1= 0#

#=>( tanx+1)^2= 0#

#=>tanx=-1#

3

Answer:

see explanation.

Explanation:

enter image source here
Let #A,B,C# be the center of circle 1, circle 2 and circle 3, respectively, as shown in the figure.
#AB=a+b, AC=a+c, BC=b+c#
In #DeltaCDB, CD^2=BC^2-BD^2#
#=> CD^2=(b+c)^2-(b-c)^2#
#=> CD^2=(b+c+b-c)*(b+c-b+c)=2b*2c=4bc#
#=> CD=sqrt(4bc)=2sqrt(bc)#
Similarly, #CF^2=(a+c)^2-(a-c)^2#
#=> CF=2sqrt(ac)#
#=> DF=CF+CD=2sqrt(ac)+2sqrt(bc)=2sqrtc(sqrta+sqrtb)#
In #DeltaAEB, BE^2=AB^2-AE^2#
#=> BE^2=(a+b)^2-(b-a)^2#
#=> BE^2=(a+b+b-a)*(a+b-b+a)=2b*2a=4ab#
#=> BE=sqrt(4ab)=2sqrt(ab)#
As #BE=DF#,
#=> 2sqrt(ab)=2sqrtc(sqrta+sqrtb)#
#=> 1/sqrtc=(sqrta+sqrtb)/sqrt(ab)#
#=> 1/sqrtc=sqrta/sqrt(ab)+sqrtb/sqrt(ab)#
#=> 1/sqrtc=1/sqrtb+1/sqrta#
#=> 1/sqrtc=1/sqrta+1/sqrtb#

1

Answer:

Hence, the orthocentre of #triangle ABC# is #C(6,3)#

Explanation:

Let , #triangle ABC # ,be the triangle with corners at

#A(2,3) ,B(6,1) and C(6,3) # .

We take, #AB=c, BC=a and CA=b#

So,

#c^2=(2-6)^2+(3-1)^2=16+4=20#

#a^2=(6-6)^2+(1-3)^2=0+4=4#

#b^2=(2-6)^2+(3-3)^2=16+0=16#

It is clear that, #a^2+b^2=4+16=20=c^2#

# i.e. color(red)(c^2=a^2+b^2=>mangleC=pi/2#

Hence, #bar(AB)# is the hypotenuse.

#:.triangle ABC # is the right angled triangle.

#:.#The orthocenter coindes with #C#

Hence, the orthocentre of #triangle ABC# is #C(6,3)#

Please see the graph:

enter image source here

2

Answer:

I made a diagram before reading about the pizza....

Explanation:

Hope it helps:

enter image source here

The green rectangle is unity (one pizza) and the little rectangles are the slices...