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1) 3x - 2/2x + 1 = 4/5 5(3x - 2) = 4(2x +1) 15x - 10 = 8x + 4 15x - 8x = 10 + 4 7x = 14 x = 14/7 = 2

2) 7 + 4x = 9x - 13 9x - 4x = 13 + 7 5x = 20 x = 20/5 = 4

2. c) 0.606006... is not rational,

3. B) (a,0) is on x - axis

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in order to prove that its on the circle, put the values x=-3 and y=0 in the equation, if a is on the circle then then the result will be zero

center C(-4,-6)

let (a,b) be the other end of diameterso,[(a-3)/2 ,(b+0)/2]= (-4,-6)so,a=-5b=-12hence other point is (-5,-12)

2)option C as it is non terminating decimal fraction3)(a,0) as a point will lie on X axis given that y is 0

Prove them congurent

yes

quadrant 2 and 4 in which sign of abscissa and ordinate are opposite

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Thepoint ofintersectionoftheaxes is calledthe origin, • Abscissa or thex-coordinate ofapointis itsdistance fromthe y-axisand the. ordinate or the y-coordinateis itsdistance fromthex-axis.

Given : PQ is parallel to BC and PQ divides triangle ABC into two parts.To find : BP/AB

Proof : InΔ APQΔ ABC,

∠ APQ =∠ ABC (As PQ is parallel to BC)

∠ PAQ =∠ BAC (Common angles)

⇒Δ APQ ~Δ ABC (BY AA similarity)

Therefore,

ar(Δ APQ)/ar(Δ ABC) = AP²/AB²

⇒ ar(Δ APQ)/2ar(Δ APQ) = AP²/AB²

⇒ 1/2 = AP²/AB²

⇒ AP/AB = 1/√2

⇒ (AB - BP)/AB = 1/√2

⇒ AB/AB - BP/AB = 1/√2

⇒ 1 - BP/AB = 1/√2

⇒ BP/AB = 1 - 1/√2

⇒ BP/AB =√2 - 1/√2

absicca is the x coordinate of the pointi) absicca=0ii) absicca=3iii) absicca=-2.

Abscissa is x coordinateordinate is y coordinatei) (6, -3) x = 6 and x = -3

ii) (-2,-9) x = -2 and y = -3

iii) (0,-12) x = 0 and y = -12

Solution:-Given : PQ is parallel to BC and PQ divides triangle ABC into two parts.To find : BP/ABProof : InΔ APQΔ ABC,∠ APQ =∠ ABC (As PQ is parallel to BC)∠ PAQ =∠ BAC (Common angles)⇒Δ APQ ~Δ ABC (BY AA similarity)Therefore,ar(Δ APQ)/ar(Δ ABC) = AP²/AB²⇒ ar(Δ APQ)/2ar(Δ APQ) = AP²/AB²⇒ 1/2 = AP²/AB²⇒ AP/AB = 1/√2⇒ (AB - BP)/AB = 1/√2⇒ AB/AB - BP/AB = 1/√2⇒ 1 - BP/AB = 1/√2⇒ BP/AB = 1 - 1/√2⇒ BP/AB =√2 - 1/√2

∆ABC And ∆ADC Both Lie On The Same Base In A Way That

ar( ∆ABC ) = ar( ∆ABD)

to prove:-∆ABC And ∆ADB Lie Between The Same Parallel.. ie:- CD || AB

construction:-

Altitude CE And DF Of ∆ACB & ∆ADB.. On AB .Now According To Question It's Said That ∆ABC And ∆ABD Both lie On The Same Base ... And Both Have

Equal Area.

Now By Construction We Have ...

CE Perpendicular To AB

And DF Perpendicular To AB

Now We Know That Lines Perpendicular To Same Line Are Parallel To Each Other Therefore ...

CE || DF --- EQ (1)

So Now It's Given That

ar ( ABC ) = ar ( ABD )

Now We Know That ..

Area of triangle=1/2*base*area

Therefore

ar(ABC) = 1/2 × AB × CE

And

ar(ABD) = 1/2 × AB × DF

Now Since Area Of ABC = Area OF ABD

Therefore We Have ...

CE=DF ---- EQ. (2)

Now In Quadrilateral CDEF

CE = DF

And

CE || DF

So We Know That If In A Quadrilateral If One Pair Of Opposite Sides Are Equal And Parallel Then The

Quadrilateral Is Parallelogram.

Therefore

CDEF Is A Parallelogram ...

Therefore

CD || EF ( opposite Sides Of Parallelogram )

That Is ..

CD||ABHence proved

I am not clear

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Triangles on the same base and between same parallels are equal in area.

Given: Two triangles ABC and DBC are on the same base BC and between same parallels EF and BC.

To Prove : ar(ΔABC) = ar(ΔDBC)

Construction : Through B, draw BE || AC intersecting line AD in E and through C draw CF || BD intersecting the line DA in F.

Proof : EACB and DFCB are parallelograms (Since two pairs of opposite sides are parallel)

Also ||gm EACB and ||gm DFCB are on the same base BC and between same parallels EF and BC.∴ ar(||gm EACB)= ar(||gm DFCB)...(i)Now AB is the diagonal of ||gm EACB∴ ar(ΔEAB) = ar(ΔABC)∴ ar(ΔABC) = (1/2)ar(||gm EACB) .....(ii)Similarly,ar(ΔDBC) = (1/2) ar(||gm DFCB)...(iii)From equations (i), (ii) and (iii), we getar(Δ ABC) = ar(Δ DBC)

Line meets x axis at point where y = 0Then,2(x+3) - 3(1+y) = 02(x + 3) - 3(1+0) = 02x + 6 - 3 = 02x = - 3x = - 3/2

Line meets x axis at (-3/2,0)

Given points are A(7,3) and B(11,0)

Then,Distance between points A and B is given by

AB^2 = (11 - 7)^2 + (0 - 3)^2AB^2 = (4)^2 + (3)^2AB^2 = 16 + 9AB = sqrt(25)AB = 5

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quadrate answer (-3 2)

Slope of line= y₂-y₁/x₂-x₁x₁, y₁=(2,5)x₂ y₂=(x, 3)slope= 2= (3-5/x-2)= 2(x-2)= -2x-2= -1x= 2-1= 1AnswerPlease like the solution

m=y2-y1/x2-x1 =3-5/1-1 =-2/0 which is not defined.

m= -2/0m= infinity is the correct answer of the given question

it will be infinite as x point is same for those pointsslope=y_2-y_1/x_2-x13-5/1-1=-2/0=infinity

-2/0 which is equal to "not defined"

First find the gradient of the line using the formulay2−y1/x2−x1

=(-1) - 3/4 - (-1)= - 4/5

Next, use the equation of a line which is

(y−y1) = m(x−x1), where m is the gradient

(y − 3) = - 4/5(x − −1)

y − 3 = - 4x/5 - 4/5

y = - 4x/5 - 4/5 + 3

y = - 4x/5 + 11/5

Therefore, Equation of line isy = - 4x/5 + 11/5

z = 70°y = 50°as DE || BC so corresponding angles

In ∆ADE70° + 50° + x = 180°x = 180° -120 = 60°

From pythagoras theroem,BD^2=15^2 - 9^2BD^2=225-81BD^2=144BD=√144BD=12mNow apply pythagoras in triangle BCDCD^2=20^2 - 12^2CD^2=400-144CD^2=256CD=√256CD=16m

If p is the midpoint then 2cp= CDso CP = 2*2.5= 5 unitPlease like the solution

Moscow is the capital of Russia.