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A = { x | 6 < x <= 12 , x is real number}

Answer D)4√2ExplanationLike if you find it useful

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BASIC PROPORTIONALITY THEOREM (BPT) : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. That is also known as Thales theorem.

GIVEN:In ∆ABC , DE || ABAD= 3x+19, DC= x+3, BE= 3x+4, EC = x

AD /DC = BE/EC

[ By Thales theorem(BPT)]

3x+19/x+3 = 3x+4/x(3x+19) × x = (3x+4) × (x+3)3x² + 19x = 3x² + 9x + 4x +123x² + 19x = 3x² + 13x +123x² - 3x² +19x -13x = 126x = 12x = 12/6= 2

x = 2

Hence, the value of x is 2 .

p(x)=4x²+3x-14x²+4x-x-14x(x+1)-1(x+1)(x+1)(4x-1)It means -1 and 1/4 are the zeroes of p(x)

hit like if you find it useful

please full ans send madi

Triangles BCD, BDA, and ABC has same height and their bases are related by CD = AD = AC/2.Hence, Area of BCD = Area of BDA = (Area of ABC)/2

Similarly, Area of ABE = Area of ACE = (Area of ABC)/2

So, Area of BCD = Area of ACE--> (Area of FDCE + Area of BEF) = (Area of FDCE + Area of AFD)--> Area of BEF = Area of AFD

Now we know, Area of BCD = Area of BDA--> (Area of FDCE + Area of BEF) = (Area of AFB + Area of AFD)--> Area of FDCE = Area of AFB

Hence, the required ratio is 1:1.

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Give me the figure.so that i will help you

where is the fig

pls give the fig it is most important

i think DE ||BC and answer is AD^2/AC^2

please ask one question at a time

tq

Thank u soon oolong much

o is the answer

1. 14 (no left over)

2. 21 (left over = 3)

My present age = 27 + 8 = 35 years

Age 8 years ago = 27 years

Then,Present age = 27 + 8= 35 years

present age=27+8=35years

Quadrant is fourthhere cosec is negativehencecosx=1/√2X=(3π/2+π/4)hencecosec(3π/2+π/4)=-cosec(π/4)=-√2

Yes it can be proved by taking determenant.

4 a²b²c² is the right answer of the following

25+25+25+25=100 cha ok by

25+25+25+25=100

Therefore,ur answer is 100.

(3/25) × 100 = 12%

i hope you need this answer

Taking square on both sides1-x^3/100=9/25x^3/100=34/25x^3=34*100/25=34*4=136x3√136

⇒ In the given figure , we can see that ⇒∠DAO = ∠BAO [Because, AB and AD are tangents in the circe]

So , we take this angls as 1 , that is , ⇒ ∠DAO = ∠BAO = 1

Also in quad. ABCD , we get, ⇒∠ABO = ∠CBO { Because, BA and BC are tangents }

⇒Also , let us take this angles as 2. that is , ⇒∠ABO = ∠CBO= 2

⇒ As same as , we can take for vertices C and as well as D.

⇒ Sum. of angles of quadrilateral ABCD = 360° { Sum of angles of quad is360°}

Therfore ,

⇒2(1 + 2 + 3 + 4) = 360° { Sum. of angles of quad is -360° } ⇒1 + 2 + 3 + 4 = 180°

Now , in Triangle AOB, ⇒∠BOA= 180 –(a + b ) ⇒ { Equation 1 }

Also , In triangle COD, ⇒∠COD = 180 – (c + d) ⇒ { Equation 2 }

⇒From Eq. 1 and 2 we get ,⇒Angle BOA + Angle COD = 360 – (a + b + c + d)

= 360° – 180°

= 180°

⇒So , we conclude that the line AB and CD subtend supplementary angles at the centreO

⇒Hence it is proved that -opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

thank u very much mam