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Answer:

Option (C)

Explanation:

enter image source here
Given family of lines #(4a+3)x - (a+1)y - (2a+1) = 0#,where #ainR#

Rearranging we get

#4ax+3x - ay-y - 2a-1 = 0#

#or,(4x- y - 2)a+(3x-y-1) = 0#

Obviously the all lines of the family mus pass through the point of intersection of two lines represented by the following two equations

#4x- y - 2=0.....[1]#

and

#3x-y-1=0........[2]

Subtracting [2] from [1] we get #x=1#

Inserting #x=1# in [1] we get #y=2#

So the coordinates of common point of intersection of all lines of the family will be #(1,2)#

Any line passing through this common point #(1,2)# may be represented as #y-2=m(x-1)....[3]#,where m is the gradient of the line , a variable one.
Now rearranging [3] in intercept form we get

#y-2=m(x-1)#

#=>mx-y=m-2#

#=>x/((m-2)/m)+y/(2-m)=1#

So area of the triangle made by this line with the positive semi axes will be given by

#A=1/2xx(m-2)/mxx(2-m)=(-m^2+4m-4)/(2m)#

#=>A=-m/2+2-2/m#

Differentiating w r, to m we get

#(dA)/(dm)=-1/2+0+2/m^2#

Imposing the condition of minimization of #A#i.e . #(dA)/(dm)=0# we get

#-1/2+2/m^2=0#

#=>m^2=4#

#=>m=-2#, as per given condition the sraight line which forms minimum area with the positive semi axes must have negative gradient.

Hence minimum area of the the triangle should be for m=-2##

#A_"min"=(-(-2)^2+4(-2)-4)/(2(-2))=4#

[It matches with Option (C)]

Answer:

# "Arc Length" = (10pi)/3 ~~ 10.47 \ m #

Explanation:

If the circle has radius #5 \ m# then the total arc length of the entire circle (ie the perimeter) is given by the standard formula:

# P_"Total" = 2 pi r #

# \ \ \ \ \ \ \ \ \ = 2 xx pi xx 5 #

# \ \ \ \ \ \ \ \ \ = 10 pi #

We require the length of the arc of a sector of #120^o#, we can represent this as a fraction of that of the entire circle:

# P_"Sector" = 120^0/360^0 xx P_"Total" #

# \ \ \ \ \ \ \ \ \ \ = 1/3 xx P_"Total" #

# \ \ \ \ \ \ \ \ \ \ = 1/3 xx 10 pi #

# \ \ \ \ \ \ \ \ \ \ = (10pi)/3 #

# \ \ \ \ \ \ \ \ \ \ ~~ 10.47 \ m #

enter image source here Let #PR=A# be the side of the isosceles triangle having coordinates of its end points as follows

#Pto(1,4)# and #Rto (5,1)#

Let the coordinates of the third point of the triangle be #(x,y)#.

As #(x,y)# is equidistant from P and R we can write

#(x-1)^2+(y-4)^2=(x-5)^2+(y-1)^2#

#=>x^2-2x+1+y^2-8y+16=x^2-10x+25+y^2-2y+1#

#=>8x-6y=9#

#=>x=(9+6y)/8......[1]#

Again #(x,y)# being equidistant from P and R, the perpendicular dropped from #(x,y)# to #PR# must bisect it, Let this foot of the perpendicular or mid point of #PR# be #T#

So coordinates of #Tto(3,2.5)#

Now height of the isosceles triangle

#H=sqrt((x-3)^2+(y-2.5)^2)#

And the base of the isosceles triangle

#PR=A=sqrt((1-5)^2+(4-1)^2)=5#

So by the problem its area

#1/2xxAxxH=15#

#=>H=30/A=30/5=6#

#sqrt((x-3)^2+(y-2.5)^2)=6#

#=>(x-3)^2+(y-2.5)^2=36....[2]#

By [2] and [1] we get

#((9+6y)/8-3)^2+(y-2.5)^2=36#

#=>1/64(6y-15)^2+(y-2.5)^2=36#

#=>(6y-15)^2+64(y-2.5)^2=36xx64#

#=>36y^2-180y+225+64y^2-320y+400=48^2#

#=>100y^2-500y+625=48^2#

#=>y^2-5y+6.25=4.8^2#

#=>(y-2.5)^2=4.8^2#

#=>y=2.5pm4.8#

So #y=7.3 and y= -2.3#

when #y=7.3#

#x=(9+6xx7.3)/8=6.6#

when #y=-2.3#

#x=(9+6xx(-2.3))/8=-0.6#

So the coordinates of third point will be

#(6.6,7.3)to"Q in figure"#

OR

#(-0.6,-2.3)to"S in figure"#

Answer:

# d = 1.496 times 10^{11} cdot 1920/3600 cdot {2 pi}/360 = 1.39254×10^9 # meters

Explanation:

Well now I have to look up what the double quote means. Probably minutes or seconds.

OK, Wikipedia says For example, 40.1875° = 40° 11' 15" .

That's 40 degrees, 11 minutes, 15 seconds, a minute being 1/60th of a degree and a second being 1/60th of a minute. (Thank you Babylonians.)

So we're to understand 1920" as

# 1920/3600 # degrees # = 1920/3600 cdot {2 pi}/360 # radians

So we're looking at a diameter

# d = 1.496 times 10^{11} cdot 1920/3600 cdot {2 pi}/360 # meters

# d = 1.39254×10^9 # meters

Check: Google #quad sqrt#

enter image source here

Please solve q 92?

dk_ch
dk_ch
Featured 4 days ago

Answer:

option (1) #6sqrt2#

Explanation:

Q92

enter image source here

Given that G is the centroid of #Delta ABC#

#GA=2sqrt3,GB=2sqrt2and GC=2#

AP median is produced such that #GP=PO#

#B,O and C,O# are joined. The quadrilateral CBCO is a parallelogram,

As #AG:GP=2:1# we get

#DeltaBGP=1/3DeltaABP=1/3*1/2DeltaABC#

Again #DeltaBGO=2DeltaBGP#

So #DeltaABC=3DeltaBGO#

Now #GO=GA=2sqrt3#

#BO=GC=2#

and #GB=2sqrt2#

So #GO^2-(2sqrt3)^2=12#

#BO^2+BG^2=2^2+(2sqrt2)^2=12#

So #Delta BGO# is rigtangled at B

So #DeltaBGO=1/2BO×BG#
Area of

#Delta ABC=3DeltaBGO=3/2xxBOxxBG=3/2xx2xx2sqrt2=6sqrt2#

Answer:

See below.

Explanation:

This question is a little ambiguous. I'm guessing you mean a volume of revolution. If this is the case then a rectangle rotated about a line could be a solid cylinder or a hollow cylinder ( Tube) if the line of rotation is parallel or perpendicular to the sides of the rectangle. If the line of rotation is neither parallel nor perpendicular then a vast number of different solids could result. A big problem here is on the definition. Strictly a prism has a base that is a polygon and faces that are flat, but you could say a cylinder has a polygonal base with an in finite number of sides and an infinite number of faces.

Here are two rotations of a rectangle around different lines.

#y=0#

enter image source here

#x=-2#

enter image source here

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